Why limits give us the exact value of the slope of the tangent line? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit?Proof that derivative of a function at a point is the slope of the tangent at the pointIs answer of limit an exact value?The slope of the tangent lineGive equation of g(t) = |t-2| / t and sketch a graph of slope of the tangent line as t varies from 0 to 4.Why find the slope other than finding min , max?How to write an equation of a line bisecting an angle in terms of the slope of the bisector.Is the definition of a tangent line flawed?How tangent line gives us a slope at one point?Definition of a Tangent line and DerivativesFind the tangent line for the following: $(operatornamearcsec x)^2$ at $x = 2$Why Is The Derivative At A Point Drawn As A Tangent Line?How can I find the slope of a line tangent to a small circle on a sphere?

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Why limits give us the exact value of the slope of the tangent line?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit?Proof that derivative of a function at a point is the slope of the tangent at the pointIs answer of limit an exact value?The slope of the tangent lineGive equation of g(t) = |t-2| / t and sketch a graph of slope of the tangent line as t varies from 0 to 4.Why find the slope other than finding min , max?How to write an equation of a line bisecting an angle in terms of the slope of the bisector.Is the definition of a tangent line flawed?How tangent line gives us a slope at one point?Definition of a Tangent line and DerivativesFind the tangent line for the following: $(operatornamearcsec x)^2$ at $x = 2$Why Is The Derivative At A Point Drawn As A Tangent Line?How can I find the slope of a line tangent to a small circle on a sphere?










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Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 days ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    2 days ago










  • $begingroup$
    Related (maybe): math.stackexchange.com/questions/2491184/…
    $endgroup$
    – Hans Lundmark
    2 days ago










  • $begingroup$
    If you think it is just an approximation, can you give some bounds on the error this approximation has?
    $endgroup$
    – Hagen von Eitzen
    2 days ago










  • $begingroup$
    Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
    $endgroup$
    – Quelklef
    2 days ago















5












$begingroup$


Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 days ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    2 days ago










  • $begingroup$
    Related (maybe): math.stackexchange.com/questions/2491184/…
    $endgroup$
    – Hans Lundmark
    2 days ago










  • $begingroup$
    If you think it is just an approximation, can you give some bounds on the error this approximation has?
    $endgroup$
    – Hagen von Eitzen
    2 days ago










  • $begingroup$
    Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
    $endgroup$
    – Quelklef
    2 days ago













5












5








5





$begingroup$


Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?










share|cite|improve this question











$endgroup$




Limits tell us how functions behave at $xto a$, not how they behave at $x = a$. However, in limits we plug $x = a$ as an approximation of $xto a$, so: why the limits give us the exact value of slope of the tangent line, despite being just an approximation of what happend around $x$?







limits tangent-line slope






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









BowPark

567720




567720










asked 2 days ago









Mohammad AlshareefMohammad Alshareef

316




316











  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 days ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    2 days ago










  • $begingroup$
    Related (maybe): math.stackexchange.com/questions/2491184/…
    $endgroup$
    – Hans Lundmark
    2 days ago










  • $begingroup$
    If you think it is just an approximation, can you give some bounds on the error this approximation has?
    $endgroup$
    – Hagen von Eitzen
    2 days ago










  • $begingroup$
    Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
    $endgroup$
    – Quelklef
    2 days ago
















  • $begingroup$
    "But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
    $endgroup$
    – Dave L. Renfro
    2 days ago











  • $begingroup$
    First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
    $endgroup$
    – Allawonder
    2 days ago










  • $begingroup$
    Related (maybe): math.stackexchange.com/questions/2491184/…
    $endgroup$
    – Hans Lundmark
    2 days ago










  • $begingroup$
    If you think it is just an approximation, can you give some bounds on the error this approximation has?
    $endgroup$
    – Hagen von Eitzen
    2 days ago










  • $begingroup$
    Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
    $endgroup$
    – Quelklef
    2 days ago















$begingroup$
"But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
$endgroup$
– Dave L. Renfro
2 days ago





$begingroup$
"But in limits we plug $x=a$ ..." --- This is only a technique that can sometimes be applied. See this answer.
$endgroup$
– Dave L. Renfro
2 days ago













$begingroup$
First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
$endgroup$
– Allawonder
2 days ago




$begingroup$
First off, we don't plug anything. That's just one heuristic technique for evaluating limits, and it works only for continuous functions. As to your question, limiting procedures give a useful way for extending concepts in a natural way (in a way that completes a class of objects) and thus help us naturally assign values to objects that don't have values in the usual sense of finite operations. In courses in analysis, it is shown that these limiting values behave essentially like their finite-procedure counterpart, and that they generalise their domain of application. These are both desirable.
$endgroup$
– Allawonder
2 days ago












$begingroup$
Related (maybe): math.stackexchange.com/questions/2491184/…
$endgroup$
– Hans Lundmark
2 days ago




$begingroup$
Related (maybe): math.stackexchange.com/questions/2491184/…
$endgroup$
– Hans Lundmark
2 days ago












$begingroup$
If you think it is just an approximation, can you give some bounds on the error this approximation has?
$endgroup$
– Hagen von Eitzen
2 days ago




$begingroup$
If you think it is just an approximation, can you give some bounds on the error this approximation has?
$endgroup$
– Hagen von Eitzen
2 days ago












$begingroup$
Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
$endgroup$
– Quelklef
2 days ago




$begingroup$
Possible duplicate (also unanswered): math.stackexchange.com/q/1053482/243259
$endgroup$
– Quelklef
2 days ago










6 Answers
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I think I understand where you're coming from. I had the same reaction! Tangent lines should exist regardless of limits and derivatives. So, when we construct the tangent line via the derivative, we should be able to check that it's correct.



I think we're both wrong—tangent lines don't rigorously exist independent of derivatives. There are too many edge cases and difficulties when you try to define them, so we either end up with 1) no definition, which gives us no way to rigorously check that the derivative is correct; or 2) we use the derivative as our definition.



To see the difficulties involved, consider $f(x)=x^3$. We'll call the tangent line $ell$, tangent at $x_0$.



We want, of course, for $ell$ to touch $f$ at $x_0$. But we want it to do so ONLY once within a small neighborhood. So perhaps we could use this as the definition of a tangent line.



However, secant lines also fit this definition! So it can't be correct. To fix this, perhaps we note that on the right side of $f$, the tangent line is always below the curve in a small neighborhood, and on the left side, it's above, and at $x_0=0$, the tangent line is just flat. Perhaps we could restrict the tangent line to abide by this—below on right, above on left, flat in middle—and have that amendment make our definition.



And that would work, I suppose. But it's an ugly piecewise definition, and, more importantly, it doesn't generalize well. To tell where the tangent line should be 'above' or 'below' uses the concept of concavity, which we need derivatives for. So we are able to rigorously define a tangent line without derivatives for $x^3$, but not in general.



The tangent line is a deceptively difficult concept to define. There isn't (as far as I know) a generally accepted definition besides derivatives that we can use to verify that derivatives give the correct results. You can always check for each individual problem you come across by eyeballing what the tangent line should be, but doing it in general is hard.






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    A line-by-line correction of OP that should answer the confusions therein:



    (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, and it coincides with the limit, then the limit in addition happens to tell us what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



    (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



    (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about bounded, nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of bounded, nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






    share|cite|improve this answer











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    • $begingroup$
      I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
      $endgroup$
      – MSalters
      2 days ago










    • $begingroup$
      "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
      $endgroup$
      – Paul Sinclair
      2 days ago










    • $begingroup$
      @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
      $endgroup$
      – Allawonder
      2 days ago










    • $begingroup$
      @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
      $endgroup$
      – Allawonder
      2 days ago


















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    $begingroup$

    I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



    I guess that what is bothering you, is the notion of a limit that you learn at first-year calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
    see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






    share|cite|improve this answer











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      The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



      Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



      enter image description here



      This is not achieved with a different slope.



      enter image description here






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        Thought I'd add something with an emphasis on geometry to the excellent answers above.



        Consider a point A on a circle and a point B elsewhere on the circle and draw a line between the two. The line touches the circle in two places. We have a secant line. Select another point B closer to A than the original B and draw another line. Again it intersects in two places, another secant line. Iterate bringing the new Points be ever closer to we get more and more secant lines until B is actually on A and the resulting line 1) intersects the circle at one point, and 2) is perpendicular to the radius through A. We have a tangent to our circle. We can associate a slope to the corresponding line.



        To many curves we can associate a tangent circle, any circle that pass through the curve at only one point. Then the slope of the tangent line to the circle is the slope of the tangent line to the curve. We have the slope of a tangent line without calculus.



        Consider the parabola $y=x^2$. Let's create a circle that intersects it at only one point.



        $(x-a)^2+(x^2-b)^2=r^2$



        $x^2+a^2-2ax+x^4+b^2-2bx^2=r^2$



        $x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$



        Under what conditions does this have only one solution on the parabola?



        We can make r arbitrarily small and find an (a,b) within r of the target point, $x_0$. So we can make the constant term of the polynomial vanish reducing it to



        $x^3+(1-2b)x-2a=0$



        We want the pont who's tangency we want to find to be a solution so we divide by $(x-x_0)$.



        We want $x_0$ to have multiplicity 3 or the quadratic to have no real solutions. This tells us the relationship between the center of our circle, the radius and the point of tangency.






        share|cite|improve this answer











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          Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
          $$ m = fracy_2-y_1x_2-x_1.$$
          If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
          $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
          If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
          $$fracf(x)-f(x_0)x-x_0$$
          when $x$ moves to $x_0$: $xrightarrow x_0$.
          Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



          Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
          $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
          and so the equation of the tangent line
          $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






          share|cite|improve this answer











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          • $begingroup$
            Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
            $endgroup$
            – Mohammad Alshareef
            2 days ago







          • 1




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            @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
            $endgroup$
            – Dave L. Renfro
            2 days ago










          • $begingroup$
            @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
            $endgroup$
            – Mohammad Alshareef
            2 days ago











          • $begingroup$
            Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
            $endgroup$
            – Dave L. Renfro
            2 days ago











          • $begingroup$
            Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
            $endgroup$
            – Mohammad Alshareef
            2 days ago












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          6 Answers
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          6 Answers
          6






          active

          oldest

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          active

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          active

          oldest

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          3












          $begingroup$

          I think I understand where you're coming from. I had the same reaction! Tangent lines should exist regardless of limits and derivatives. So, when we construct the tangent line via the derivative, we should be able to check that it's correct.



          I think we're both wrong—tangent lines don't rigorously exist independent of derivatives. There are too many edge cases and difficulties when you try to define them, so we either end up with 1) no definition, which gives us no way to rigorously check that the derivative is correct; or 2) we use the derivative as our definition.



          To see the difficulties involved, consider $f(x)=x^3$. We'll call the tangent line $ell$, tangent at $x_0$.



          We want, of course, for $ell$ to touch $f$ at $x_0$. But we want it to do so ONLY once within a small neighborhood. So perhaps we could use this as the definition of a tangent line.



          However, secant lines also fit this definition! So it can't be correct. To fix this, perhaps we note that on the right side of $f$, the tangent line is always below the curve in a small neighborhood, and on the left side, it's above, and at $x_0=0$, the tangent line is just flat. Perhaps we could restrict the tangent line to abide by this—below on right, above on left, flat in middle—and have that amendment make our definition.



          And that would work, I suppose. But it's an ugly piecewise definition, and, more importantly, it doesn't generalize well. To tell where the tangent line should be 'above' or 'below' uses the concept of concavity, which we need derivatives for. So we are able to rigorously define a tangent line without derivatives for $x^3$, but not in general.



          The tangent line is a deceptively difficult concept to define. There isn't (as far as I know) a generally accepted definition besides derivatives that we can use to verify that derivatives give the correct results. You can always check for each individual problem you come across by eyeballing what the tangent line should be, but doing it in general is hard.






          share|cite|improve this answer











          $endgroup$

















            3












            $begingroup$

            I think I understand where you're coming from. I had the same reaction! Tangent lines should exist regardless of limits and derivatives. So, when we construct the tangent line via the derivative, we should be able to check that it's correct.



            I think we're both wrong—tangent lines don't rigorously exist independent of derivatives. There are too many edge cases and difficulties when you try to define them, so we either end up with 1) no definition, which gives us no way to rigorously check that the derivative is correct; or 2) we use the derivative as our definition.



            To see the difficulties involved, consider $f(x)=x^3$. We'll call the tangent line $ell$, tangent at $x_0$.



            We want, of course, for $ell$ to touch $f$ at $x_0$. But we want it to do so ONLY once within a small neighborhood. So perhaps we could use this as the definition of a tangent line.



            However, secant lines also fit this definition! So it can't be correct. To fix this, perhaps we note that on the right side of $f$, the tangent line is always below the curve in a small neighborhood, and on the left side, it's above, and at $x_0=0$, the tangent line is just flat. Perhaps we could restrict the tangent line to abide by this—below on right, above on left, flat in middle—and have that amendment make our definition.



            And that would work, I suppose. But it's an ugly piecewise definition, and, more importantly, it doesn't generalize well. To tell where the tangent line should be 'above' or 'below' uses the concept of concavity, which we need derivatives for. So we are able to rigorously define a tangent line without derivatives for $x^3$, but not in general.



            The tangent line is a deceptively difficult concept to define. There isn't (as far as I know) a generally accepted definition besides derivatives that we can use to verify that derivatives give the correct results. You can always check for each individual problem you come across by eyeballing what the tangent line should be, but doing it in general is hard.






            share|cite|improve this answer











            $endgroup$















              3












              3








              3





              $begingroup$

              I think I understand where you're coming from. I had the same reaction! Tangent lines should exist regardless of limits and derivatives. So, when we construct the tangent line via the derivative, we should be able to check that it's correct.



              I think we're both wrong—tangent lines don't rigorously exist independent of derivatives. There are too many edge cases and difficulties when you try to define them, so we either end up with 1) no definition, which gives us no way to rigorously check that the derivative is correct; or 2) we use the derivative as our definition.



              To see the difficulties involved, consider $f(x)=x^3$. We'll call the tangent line $ell$, tangent at $x_0$.



              We want, of course, for $ell$ to touch $f$ at $x_0$. But we want it to do so ONLY once within a small neighborhood. So perhaps we could use this as the definition of a tangent line.



              However, secant lines also fit this definition! So it can't be correct. To fix this, perhaps we note that on the right side of $f$, the tangent line is always below the curve in a small neighborhood, and on the left side, it's above, and at $x_0=0$, the tangent line is just flat. Perhaps we could restrict the tangent line to abide by this—below on right, above on left, flat in middle—and have that amendment make our definition.



              And that would work, I suppose. But it's an ugly piecewise definition, and, more importantly, it doesn't generalize well. To tell where the tangent line should be 'above' or 'below' uses the concept of concavity, which we need derivatives for. So we are able to rigorously define a tangent line without derivatives for $x^3$, but not in general.



              The tangent line is a deceptively difficult concept to define. There isn't (as far as I know) a generally accepted definition besides derivatives that we can use to verify that derivatives give the correct results. You can always check for each individual problem you come across by eyeballing what the tangent line should be, but doing it in general is hard.






              share|cite|improve this answer











              $endgroup$



              I think I understand where you're coming from. I had the same reaction! Tangent lines should exist regardless of limits and derivatives. So, when we construct the tangent line via the derivative, we should be able to check that it's correct.



              I think we're both wrong—tangent lines don't rigorously exist independent of derivatives. There are too many edge cases and difficulties when you try to define them, so we either end up with 1) no definition, which gives us no way to rigorously check that the derivative is correct; or 2) we use the derivative as our definition.



              To see the difficulties involved, consider $f(x)=x^3$. We'll call the tangent line $ell$, tangent at $x_0$.



              We want, of course, for $ell$ to touch $f$ at $x_0$. But we want it to do so ONLY once within a small neighborhood. So perhaps we could use this as the definition of a tangent line.



              However, secant lines also fit this definition! So it can't be correct. To fix this, perhaps we note that on the right side of $f$, the tangent line is always below the curve in a small neighborhood, and on the left side, it's above, and at $x_0=0$, the tangent line is just flat. Perhaps we could restrict the tangent line to abide by this—below on right, above on left, flat in middle—and have that amendment make our definition.



              And that would work, I suppose. But it's an ugly piecewise definition, and, more importantly, it doesn't generalize well. To tell where the tangent line should be 'above' or 'below' uses the concept of concavity, which we need derivatives for. So we are able to rigorously define a tangent line without derivatives for $x^3$, but not in general.



              The tangent line is a deceptively difficult concept to define. There isn't (as far as I know) a generally accepted definition besides derivatives that we can use to verify that derivatives give the correct results. You can always check for each individual problem you come across by eyeballing what the tangent line should be, but doing it in general is hard.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 days ago

























              answered 2 days ago









              QuelklefQuelklef

              465311




              465311





















                  3












                  $begingroup$

                  A line-by-line correction of OP that should answer the confusions therein:



                  (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, and it coincides with the limit, then the limit in addition happens to tell us what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                  (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                  (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about bounded, nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of bounded, nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                  share|cite|improve this answer











                  $endgroup$












                  • $begingroup$
                    I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                    $endgroup$
                    – MSalters
                    2 days ago










                  • $begingroup$
                    "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                    $endgroup$
                    – Paul Sinclair
                    2 days ago










                  • $begingroup$
                    @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                    $endgroup$
                    – Allawonder
                    2 days ago










                  • $begingroup$
                    @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                    $endgroup$
                    – Allawonder
                    2 days ago















                  3












                  $begingroup$

                  A line-by-line correction of OP that should answer the confusions therein:



                  (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, and it coincides with the limit, then the limit in addition happens to tell us what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                  (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                  (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about bounded, nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of bounded, nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                  share|cite|improve this answer











                  $endgroup$












                  • $begingroup$
                    I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                    $endgroup$
                    – MSalters
                    2 days ago










                  • $begingroup$
                    "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                    $endgroup$
                    – Paul Sinclair
                    2 days ago










                  • $begingroup$
                    @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                    $endgroup$
                    – Allawonder
                    2 days ago










                  • $begingroup$
                    @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                    $endgroup$
                    – Allawonder
                    2 days ago













                  3












                  3








                  3





                  $begingroup$

                  A line-by-line correction of OP that should answer the confusions therein:



                  (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, and it coincides with the limit, then the limit in addition happens to tell us what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                  (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                  (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about bounded, nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of bounded, nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.






                  share|cite|improve this answer











                  $endgroup$



                  A line-by-line correction of OP that should answer the confusions therein:



                  (1) Limits tell us how a function behaves nearby $x=a.$ The function doesn't even have to be defined at $a.$ If it is also defined there, and it coincides with the limit, then the limit in addition happens to tell us what's happening at $a,$ contrary to your first sentence. But the important thing to note is that limits tell us how a function is behaving in the neighborhood of some point in its domain.



                  (2) From above, it follows that $f(a)$ need not exist at all in order for $lim_xto af(x)$ to exist. Indeed, even if $f(a)$ exists, it need not coincide with the limiting value. In essence, limiting values have nothing essentially to do with values. But in cases where $lim_xto af(x)$ exists and coincides with $f(a),$ we may use this fact to evaluate the limiting value, but it is no way an approximation. There's nothing about approximations here. The conditions are all equalities. Again, the salient point here is that we can't always evaluate limits, when they exist, by a substitution as you think. This only applies to functions continuous at the point in question.



                  (3) You have yet to understand the concept of limit properly. It is not an approximation to anything, but rather that towards which we approximate; thus by definition it will sometimes transcend every approximant. Rather, limits are a way to extend usual concepts into wider domains of applicability in such a way that the new objects properly generalise the old -- that is, they possess all important properties of the old objects, and contain the latter as a subclass. In particular here, bringing in limiting operations helps us to assign a definite meaning to some sequences of approximations that distinguish themselves from others. Note that whereas some terms of these types of sequence approximate the limiting value, this value itself is decidedly not an approximation, but a uniquely defined value, fixed by this sequence (and an assumption about bounded, nonempty subsets of real numbers that genuinely and smoothly extends the concept of maximum or minimum values of bounded, nonempty subsets of numbers). This well-defined limiting value, whenever it exists, has been shown many times over to legitimately (in a sense that can be made precise) extend the domain of application of the concept under study. Here, it follows that whenever a sequence of approximating slopes converges, we can choose to call (by a justified extension) the limit a slope too, indeed the slope, defined by that sequence. That's simply what it means to work with real numbers. If you've not understood this property (technically called completeness) then you've not begun to understand the real numbers.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  AllawonderAllawonder

                  2,359617




                  2,359617











                  • $begingroup$
                    I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                    $endgroup$
                    – MSalters
                    2 days ago










                  • $begingroup$
                    "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                    $endgroup$
                    – Paul Sinclair
                    2 days ago










                  • $begingroup$
                    @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                    $endgroup$
                    – Allawonder
                    2 days ago










                  • $begingroup$
                    @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                    $endgroup$
                    – Allawonder
                    2 days ago
















                  • $begingroup$
                    I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                    $endgroup$
                    – MSalters
                    2 days ago










                  • $begingroup$
                    "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                    $endgroup$
                    – Paul Sinclair
                    2 days ago










                  • $begingroup$
                    @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                    $endgroup$
                    – Allawonder
                    2 days ago










                  • $begingroup$
                    @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                    $endgroup$
                    – Allawonder
                    2 days ago















                  $begingroup$
                  I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                  $endgroup$
                  – MSalters
                  2 days ago




                  $begingroup$
                  I think the "tangent line" in the title is a hint that the question should have been answered at a simpler level. Yes, there's no need for f(a) to exist, or to equal the limit. But students which are still working with the notion of tangent lines are unlikely to be dealing with discontinuities, I think.
                  $endgroup$
                  – MSalters
                  2 days ago












                  $begingroup$
                  "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                  $endgroup$
                  – Paul Sinclair
                  2 days ago




                  $begingroup$
                  "If it is also defined there, then the limit tells us in addition what's happening at $a$" - The limit itself never tells us what is happening at $a$. If $f$ is continuous at $a$, then the limit is indeed what is happening at $a$, but it was the condition of continuity that told us this, not the limit.
                  $endgroup$
                  – Paul Sinclair
                  2 days ago












                  $begingroup$
                  @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                  $endgroup$
                  – Allawonder
                  2 days ago




                  $begingroup$
                  @PaulSinclair I would agree that the limit never tells what's happening at $a.$ As I emphasized, the main point is that the limit tells what's happening close to $a.$ However, I think it was something like what's in your second paragraph I was trying to express. But I think the emphases I made should point the way. In any case I've edited the relevant part. Thanks.
                  $endgroup$
                  – Allawonder
                  2 days ago












                  $begingroup$
                  @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                  $endgroup$
                  – Allawonder
                  2 days ago




                  $begingroup$
                  @MSalters I don't understand what you mean by simpler level. My emphases couldn't have been simpler. What I noticed was the erroneous notion of limiting procedures in OP, which I think was the source of the confusions, and this I attempted to set right. Once OP understands that whenever a sequence converges, then its limit is unique, and they also see that limits are a way to complete $mathbf Q,$ and they also agree to call the limit of a sequence of secant slopes a tangent slope, then everything falls into place.
                  $endgroup$
                  – Allawonder
                  2 days ago











                  2












                  $begingroup$

                  I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                  I guess that what is bothering you, is the notion of a limit that you learn at first-year calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                  see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                  share|cite|improve this answer











                  $endgroup$

















                    2












                    $begingroup$

                    I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                    I guess that what is bothering you, is the notion of a limit that you learn at first-year calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                    see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                    share|cite|improve this answer











                    $endgroup$















                      2












                      2








                      2





                      $begingroup$

                      I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                      I guess that what is bothering you, is the notion of a limit that you learn at first-year calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                      see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit






                      share|cite|improve this answer











                      $endgroup$



                      I think the problem with your thinking is that, you are assuming that there is something that is predefined by "god" which is the slope at a point that has exact value, and we are trying to reach it by the processes of taking a limit, and we can never reach that value. But that is not the case. What we do, is that we define the notion of the slope at a point , by the processes of taking a limit. And what ever the exact value that comes out of the limit, is what we call the slope at that point.



                      I guess that what is bothering you, is the notion of a limit that you learn at first-year calculus , and i felt the same when i first learn it, because at some stage of the problem we say that x approaches some value a , then at another stage we substitute a for x. But to understand what is actually going on, you have to understand the epsilon-delta definition of a limit.
                      see:https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      yousef magablehyousef magableh

                      10810




                      10810





















                          1












                          $begingroup$

                          The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



                          Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



                          enter image description here



                          This is not achieved with a different slope.



                          enter image description here






                          share|cite|improve this answer











                          $endgroup$

















                            1












                            $begingroup$

                            The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



                            Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



                            enter image description here



                            This is not achieved with a different slope.



                            enter image description here






                            share|cite|improve this answer











                            $endgroup$















                              1












                              1








                              1





                              $begingroup$

                              The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



                              Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



                              enter image description here



                              This is not achieved with a different slope.



                              enter image description here






                              share|cite|improve this answer











                              $endgroup$



                              The idea is that the approximation gets better and better as $x$ gets closer to $a$, and, so to say, at $x=a$ it is no more approximative.



                              Below, a function in blue, the tangent in green and the difference in magenta. As you see, the difference decreases very quickly close to the tangency point.



                              enter image description here



                              This is not achieved with a different slope.



                              enter image description here







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 2 days ago

























                              answered 2 days ago









                              Yves DaoustYves Daoust

                              133k676232




                              133k676232





















                                  1












                                  $begingroup$

                                  Thought I'd add something with an emphasis on geometry to the excellent answers above.



                                  Consider a point A on a circle and a point B elsewhere on the circle and draw a line between the two. The line touches the circle in two places. We have a secant line. Select another point B closer to A than the original B and draw another line. Again it intersects in two places, another secant line. Iterate bringing the new Points be ever closer to we get more and more secant lines until B is actually on A and the resulting line 1) intersects the circle at one point, and 2) is perpendicular to the radius through A. We have a tangent to our circle. We can associate a slope to the corresponding line.



                                  To many curves we can associate a tangent circle, any circle that pass through the curve at only one point. Then the slope of the tangent line to the circle is the slope of the tangent line to the curve. We have the slope of a tangent line without calculus.



                                  Consider the parabola $y=x^2$. Let's create a circle that intersects it at only one point.



                                  $(x-a)^2+(x^2-b)^2=r^2$



                                  $x^2+a^2-2ax+x^4+b^2-2bx^2=r^2$



                                  $x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$



                                  Under what conditions does this have only one solution on the parabola?



                                  We can make r arbitrarily small and find an (a,b) within r of the target point, $x_0$. So we can make the constant term of the polynomial vanish reducing it to



                                  $x^3+(1-2b)x-2a=0$



                                  We want the pont who's tangency we want to find to be a solution so we divide by $(x-x_0)$.



                                  We want $x_0$ to have multiplicity 3 or the quadratic to have no real solutions. This tells us the relationship between the center of our circle, the radius and the point of tangency.






                                  share|cite|improve this answer











                                  $endgroup$

















                                    1












                                    $begingroup$

                                    Thought I'd add something with an emphasis on geometry to the excellent answers above.



                                    Consider a point A on a circle and a point B elsewhere on the circle and draw a line between the two. The line touches the circle in two places. We have a secant line. Select another point B closer to A than the original B and draw another line. Again it intersects in two places, another secant line. Iterate bringing the new Points be ever closer to we get more and more secant lines until B is actually on A and the resulting line 1) intersects the circle at one point, and 2) is perpendicular to the radius through A. We have a tangent to our circle. We can associate a slope to the corresponding line.



                                    To many curves we can associate a tangent circle, any circle that pass through the curve at only one point. Then the slope of the tangent line to the circle is the slope of the tangent line to the curve. We have the slope of a tangent line without calculus.



                                    Consider the parabola $y=x^2$. Let's create a circle that intersects it at only one point.



                                    $(x-a)^2+(x^2-b)^2=r^2$



                                    $x^2+a^2-2ax+x^4+b^2-2bx^2=r^2$



                                    $x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$



                                    Under what conditions does this have only one solution on the parabola?



                                    We can make r arbitrarily small and find an (a,b) within r of the target point, $x_0$. So we can make the constant term of the polynomial vanish reducing it to



                                    $x^3+(1-2b)x-2a=0$



                                    We want the pont who's tangency we want to find to be a solution so we divide by $(x-x_0)$.



                                    We want $x_0$ to have multiplicity 3 or the quadratic to have no real solutions. This tells us the relationship between the center of our circle, the radius and the point of tangency.






                                    share|cite|improve this answer











                                    $endgroup$















                                      1












                                      1








                                      1





                                      $begingroup$

                                      Thought I'd add something with an emphasis on geometry to the excellent answers above.



                                      Consider a point A on a circle and a point B elsewhere on the circle and draw a line between the two. The line touches the circle in two places. We have a secant line. Select another point B closer to A than the original B and draw another line. Again it intersects in two places, another secant line. Iterate bringing the new Points be ever closer to we get more and more secant lines until B is actually on A and the resulting line 1) intersects the circle at one point, and 2) is perpendicular to the radius through A. We have a tangent to our circle. We can associate a slope to the corresponding line.



                                      To many curves we can associate a tangent circle, any circle that pass through the curve at only one point. Then the slope of the tangent line to the circle is the slope of the tangent line to the curve. We have the slope of a tangent line without calculus.



                                      Consider the parabola $y=x^2$. Let's create a circle that intersects it at only one point.



                                      $(x-a)^2+(x^2-b)^2=r^2$



                                      $x^2+a^2-2ax+x^4+b^2-2bx^2=r^2$



                                      $x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$



                                      Under what conditions does this have only one solution on the parabola?



                                      We can make r arbitrarily small and find an (a,b) within r of the target point, $x_0$. So we can make the constant term of the polynomial vanish reducing it to



                                      $x^3+(1-2b)x-2a=0$



                                      We want the pont who's tangency we want to find to be a solution so we divide by $(x-x_0)$.



                                      We want $x_0$ to have multiplicity 3 or the quadratic to have no real solutions. This tells us the relationship between the center of our circle, the radius and the point of tangency.






                                      share|cite|improve this answer











                                      $endgroup$



                                      Thought I'd add something with an emphasis on geometry to the excellent answers above.



                                      Consider a point A on a circle and a point B elsewhere on the circle and draw a line between the two. The line touches the circle in two places. We have a secant line. Select another point B closer to A than the original B and draw another line. Again it intersects in two places, another secant line. Iterate bringing the new Points be ever closer to we get more and more secant lines until B is actually on A and the resulting line 1) intersects the circle at one point, and 2) is perpendicular to the radius through A. We have a tangent to our circle. We can associate a slope to the corresponding line.



                                      To many curves we can associate a tangent circle, any circle that pass through the curve at only one point. Then the slope of the tangent line to the circle is the slope of the tangent line to the curve. We have the slope of a tangent line without calculus.



                                      Consider the parabola $y=x^2$. Let's create a circle that intersects it at only one point.



                                      $(x-a)^2+(x^2-b)^2=r^2$



                                      $x^2+a^2-2ax+x^4+b^2-2bx^2=r^2$



                                      $x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$



                                      Under what conditions does this have only one solution on the parabola?



                                      We can make r arbitrarily small and find an (a,b) within r of the target point, $x_0$. So we can make the constant term of the polynomial vanish reducing it to



                                      $x^3+(1-2b)x-2a=0$



                                      We want the pont who's tangency we want to find to be a solution so we divide by $(x-x_0)$.



                                      We want $x_0$ to have multiplicity 3 or the quadratic to have no real solutions. This tells us the relationship between the center of our circle, the radius and the point of tangency.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 2 days ago

























                                      answered 2 days ago









                                      TurlocTheRedTurlocTheRed

                                      1,034411




                                      1,034411





















                                          0












                                          $begingroup$

                                          Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                                          $$ m = fracy_2-y_1x_2-x_1.$$
                                          If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                                          $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                                          If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                                          $$fracf(x)-f(x_0)x-x_0$$
                                          when $x$ moves to $x_0$: $xrightarrow x_0$.
                                          Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                                          Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                                          $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                                          and so the equation of the tangent line
                                          $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                                          share|cite|improve this answer











                                          $endgroup$












                                          • $begingroup$
                                            Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago







                                          • 1




                                            $begingroup$
                                            @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago










                                          • $begingroup$
                                            @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago











                                          • $begingroup$
                                            Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago











                                          • $begingroup$
                                            Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago
















                                          0












                                          $begingroup$

                                          Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                                          $$ m = fracy_2-y_1x_2-x_1.$$
                                          If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                                          $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                                          If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                                          $$fracf(x)-f(x_0)x-x_0$$
                                          when $x$ moves to $x_0$: $xrightarrow x_0$.
                                          Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                                          Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                                          $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                                          and so the equation of the tangent line
                                          $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                                          share|cite|improve this answer











                                          $endgroup$












                                          • $begingroup$
                                            Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago







                                          • 1




                                            $begingroup$
                                            @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago










                                          • $begingroup$
                                            @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago











                                          • $begingroup$
                                            Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago











                                          • $begingroup$
                                            Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago














                                          0












                                          0








                                          0





                                          $begingroup$

                                          Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                                          $$ m = fracy_2-y_1x_2-x_1.$$
                                          If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                                          $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                                          If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                                          $$fracf(x)-f(x_0)x-x_0$$
                                          when $x$ moves to $x_0$: $xrightarrow x_0$.
                                          Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                                          Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                                          $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                                          and so the equation of the tangent line
                                          $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$






                                          share|cite|improve this answer











                                          $endgroup$



                                          Well, if you have two points $(x_1,y_1)$ and $(x_2,y_2)$ in the plane, then the slope of the unique line through these points is
                                          $$ m = fracy_2-y_1x_2-x_1.$$
                                          If you consider a (differentiable) function $f:Bbb RrightarrowBbb R$, then the slope of the line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is
                                          $$ m = fracf(x_2)-f(x_1)x_2-x_1.$$
                                          If you take the points $x_0$ and $x$ at the x-axis, then consider the quotient
                                          $$fracf(x)-f(x_0)x-x_0$$
                                          when $x$ moves to $x_0$: $xrightarrow x_0$.
                                          Then this quotient becomes (in the limit) the slope of the tangent line of $f$ at the point $x_0$. This is particularly easy to understand if you draw a picture.



                                          Denote the slope in the limit as $f'(x_0)$. Then the above quotient gives
                                          $$f'(x_0) = fracf(x)-f(x_0)x-x_0$$
                                          and so the equation of the tangent line
                                          $$f(x) = f'(x_0)(x-x_0) + f(x_0).$$







                                          share|cite|improve this answer














                                          share|cite|improve this answer



                                          share|cite|improve this answer








                                          edited 2 days ago

























                                          answered 2 days ago









                                          WuestenfuxWuestenfux

                                          5,5911513




                                          5,5911513











                                          • $begingroup$
                                            Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago







                                          • 1




                                            $begingroup$
                                            @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago










                                          • $begingroup$
                                            @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago











                                          • $begingroup$
                                            Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago











                                          • $begingroup$
                                            Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago

















                                          • $begingroup$
                                            Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago







                                          • 1




                                            $begingroup$
                                            @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago










                                          • $begingroup$
                                            @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago











                                          • $begingroup$
                                            Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                            $endgroup$
                                            – Dave L. Renfro
                                            2 days ago











                                          • $begingroup$
                                            Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                            $endgroup$
                                            – Mohammad Alshareef
                                            2 days ago
















                                          $begingroup$
                                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago





                                          $begingroup$
                                          Since $x_0$ is an approximate , Then why we get the exact value of the slope ?
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago





                                          1




                                          1




                                          $begingroup$
                                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                          $endgroup$
                                          – Dave L. Renfro
                                          2 days ago




                                          $begingroup$
                                          @Mohammad Alshareef: The "exact value of the slope at a given point" is defined using a limit, or do you have another way of defining it? Your question seems to me like asking why we use three letters to write the word "big". (Answer: Because the word "big" is defined by making use of three letters, or something to this effect.)
                                          $endgroup$
                                          – Dave L. Renfro
                                          2 days ago












                                          $begingroup$
                                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago





                                          $begingroup$
                                          @DaveL.Renfro : Does $f(xto a) = f(x=a)$ ?
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago













                                          $begingroup$
                                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                          $endgroup$
                                          – Dave L. Renfro
                                          2 days ago





                                          $begingroup$
                                          Yes, but that's because (i) the function $g(x) = x$ is continuous [note that your two inputs are "$x rightarrow a$" and "$a$"; these are equal because of (i)] and (ii) $f$ is (presumably) a function [$f$ a function implies that whenever $c=d,$ then we have $f(c) = f(d)].$ Perhaps you wanted to say "limit as $x rightarrow a$ of $f(x)$" is equal to "$f(a)$"? (This is not true, however, unless $f$ is continuous at $x=0.)$
                                          $endgroup$
                                          – Dave L. Renfro
                                          2 days ago













                                          $begingroup$
                                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago





                                          $begingroup$
                                          Actually ,This explain everything ,But I am not satisfied , Since $(xto a) + h = a$ , $h ≠ 0$ , Then $f(xto a)$ should not be equal to $f(a)$.
                                          $endgroup$
                                          – Mohammad Alshareef
                                          2 days ago


















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