Limit to 0 ambiguity [on hold] The 2019 Stack Overflow Developer Survey Results Are InNeed help with a limitHow to evaluate indeterminate form of a limitThe limit of $ lnx + cotx$Limit of a rational function to the power of xEvaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Evaluate the following limit without L'HopitalLimit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists
Why are there uneven bright areas in this photo of black hole?
I am an eight letter word. What am I?
Straighten subgroup lattice
Did any laptop computers have a built-in 5 1/4 inch floppy drive?
If a sorcerer casts the Banishment spell on a PC while in Avernus, does the PC return to their home plane?
Falsification in Math vs Science
$EDITOR environment variable won't set
What could be the right powersource for 15 seconds lifespan disposable giant chainsaw?
Is it safe to harvest rainwater that fell on solar panels?
Why can't devices on different VLANs, but on the same subnet, communicate?
What do these terms in Caesar's Gallic wars mean?
Why couldn't they take pictures of a closer black hole?
Why not take a picture of a closer black hole?
Keeping a retro style to sci-fi spaceships?
Worn-tile Scrabble
Get name of standard action overriden in Visualforce contorller
Why isn't the circumferential light around the M87 black hole's event horizon symmetric?
What is the most efficient way to store a numeric range?
Button changing its text & action. Good or terrible?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
Will it cause any balance problems to have PCs level up and gain the benefits of a long rest mid-fight?
How to support a colleague who finds meetings extremely tiring?
Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?
How do I free up internal storage if I don't have any apps downloaded?
Limit to 0 ambiguity [on hold]
The 2019 Stack Overflow Developer Survey Results Are InNeed help with a limitHow to evaluate indeterminate form of a limitThe limit of $ lnx + cotx$Limit of a rational function to the power of xEvaluating the limit of $lim_xtoinfty(sqrtfracx^3x+2-x)$.Evaluate the following limit without L'HopitalLimit of: $ -x+sqrtx^2+x $ for $ xtoinfty $Limit with integral and powerWhat is the result of the following limit?Determining if a multivariable limit exists
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited yesterday
user8718165
1167
asked yesterday
J.MohJ.Moh
535
$endgroup$
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited yesterday
user8718165
1167
asked yesterday
J.MohJ.Moh
535
$endgroup$
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday
add a comment |
$begingroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
edited yesterday
user8718165
1167
asked yesterday
J.MohJ.Moh
535
$endgroup$
I can't determine the limit of such form:
$$lim_x to 0 frac1x, $$
$$+infty~textor -infty$$
I tried to get around it, no success.
limits
limits
edited yesterday
user8718165
1167
asked yesterday
J.MohJ.Moh
535
edited yesterday
user8718165
1167
asked yesterday
J.MohJ.Moh
535
edited yesterday
user8718165
1167
edited yesterday
user8718165
1167
edited yesterday
user8718165
1167
1167
asked yesterday
J.MohJ.Moh
535
asked yesterday
J.MohJ.Moh
535
asked yesterday
J.MohJ.Moh
535
535
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
put on hold as off-topic by RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Saad, Lord Shark the Unknown, Jyrki Lahtonen, Cesareo
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday
1
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
3
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered yesterday
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:
As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
answered yesterday
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered yesterday
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered yesterday
MPWMPW
31.2k12157
$endgroup$
add a comment |
$begingroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered yesterday
MPWMPW
31.2k12157
$endgroup$
The limit does not exist (even allowing for an infinite limit, which some definitions may not allow) since it depends on the direction of approach, as you have observed.
answered yesterday
MPWMPW
31.2k12157
answered yesterday
MPWMPW
31.2k12157
answered yesterday
MPWMPW
31.2k12157
answered yesterday
MPWMPW
31.2k12157
31.2k12157
add a comment |
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:
As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
answered yesterday
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:
As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
answered yesterday
Michael RybkinMichael Rybkin
4,264422
$endgroup$
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
add a comment |
$begingroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:
As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
answered yesterday
Michael RybkinMichael Rybkin
4,264422
$endgroup$
It's instructive to take a look at the graph of $f(x)=frac1x$ to better see what exactly is going on with the function as $x$ goes to zero:
As you can probably guess, this limit should be split into two one-sided limits for a proper analysis because the function behaves differently depending on which side you approach the value of zero from. As $x$ approaches $0$ from the left (denoted by $xto 0^-$), the function grows without bound negatively. As $x$ approaches $0$ from the right (denoted by $xto 0^+$), the function grows without bound positively. Analytically, these two facts are written as follows:
$$lim_xto 0^-frac1x=-infty, lim_xto 0^+frac1x=+infty.$$
For a limit to exist as $x$ approaches a particular point, the two one-sided limits at that point must be equal. Apparently, $lim_xto 0^-frac1xnelim_xto 0^+frac1x$. Thus, $lim_xto 0frac1x=DNE$ (does not exist).
Strictly speaking, infinite limits are also considered limits that do not exist (a limit that exists should be a number and infinity is not a number). Nevertheless, we still write the equality sign and denote what kind of infinity the function is going to. This helps us better understand the behavior of the function. For example, $lim_xto 2g(x)=-infty$ means that as $x$ approaches $2$ from both sides (from the left and from the right), the function $g(x)$ keeps growing without bound negatively. "It goes to negative infinity" is a simpler way to put it. And this is valuable information because it tells us something about the behavior of the function. It's better to know what kind of infinity a function is going off to than just stating the fact that it simply does not exist.
answered yesterday
Michael RybkinMichael Rybkin
4,264422
edited yesterday
answered yesterday
Michael RybkinMichael Rybkin
4,264422
answered yesterday
Michael RybkinMichael Rybkin
4,264422
answered yesterday
Michael RybkinMichael Rybkin
4,264422
4,264422
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
add a comment |
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
$begingroup$
Thank you so much!!!
$endgroup$
– J.Moh
yesterday
add a comment |
1
$begingroup$
Simply: the limit does not exist. You can say that $$lim_xto 0^+ frac1x=infty$$ and $$lim_xto 0^- frac1x=-infty.$$
$endgroup$
– Dave
yesterday
3
$begingroup$
How are you defining a limit?
$endgroup$
– John Doe
yesterday
$begingroup$
Just to 0, that s what I am asking for, as it is clear, it doesn t exist.
$endgroup$
– J.Moh
yesterday