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Clarify a step in the proof that if $limfracf(x)g(x)$ exists, then $lim(1+f(x))^1/g(x)=e^lim f(x)/g(x)$

Finding $lim limits_x to 0 frac1 - cos xx$, given $lim limits_x to 0 fracsin xx = 1$Calculating the limit: $lim limits_x to 0$ $fracln(fracsin xx)x^2. $Spot mistake in finding $lim limits_xto1left(frac x x-1 - frac1 log(x) right)$Proving the surprising limit: $limlimits_n to 0 fracx^n-y^nn$ $=$ log$fracxy$Limit of this expression $ lim limits_nto infty left( frac15n+7n logleft( nright) right) $Question about prove when $x_n$ and $y_n$ convergent $lim x_n = lim y_n$ where $x_n+1=fracx_n+y_n2$, $y_n+1=frac2x_ny_nx_n+y_n$Calculate $limlimits_n to infty frac1ncdotlogleft(3^fracn1 + 3^fracn2 + dots + 3^fracnnright)$Proving that $limlimits_h to 0frach$ does not exist at $x=0$How can I solve $limlimits_n to infty fracn log_2 log_2 n3^log_2 n^2= 0$Is this way of finding $limlimits_xto +infty(x-ln(x^2+1))$ valid?

1

$begingroup$

There is a theorem which says "If $limlimits_xto a f(x) =limlimits_xto a g(x)=0$ such that $limlimits_xto a fracf(x)g(x)$ exists, then $limlimits_x to a 1+f(x)^frac1g(x) = e^limlimits_xto a fracf(x)g(x)$

The proof for this theorem goes as such:
$$textLet A=lim_xto a 1+f(x)^1over g(x).text Then, $$
$$log_e A = lim_xto a fraclog1+f(x)g(x)$$
$$implies log_e A = lim_xto a fraclog1+f(x)f(x) cdot fracf(x)g(x)$$
$$implies log_e A = lim_xto a fracf(x)g(x)$$
$$implies A=e^limlimits_xto a fracf(x)g(x)$$

Now my doubt here is... how did the first step become second step by taking log on both sides? How did the log go inside the limit? That is my confusion.

That is, $A=lim_xto a 1+f(x)^1over g(x)$ became $log_e A = lim_xto a fraclog1+f(x)g(x)$. How?

Hope you get what my doubt is. Thanks.

calculus limits

share|cite|improve this question

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

$endgroup$

add a comment | 

1

$begingroup$

There is a theorem which says "If $limlimits_xto a f(x) =limlimits_xto a g(x)=0$ such that $limlimits_xto a fracf(x)g(x)$ exists, then $limlimits_x to a 1+f(x)^frac1g(x) = e^limlimits_xto a fracf(x)g(x)$

The proof for this theorem goes as such:
$$textLet A=lim_xto a 1+f(x)^1over g(x).text Then, $$
$$log_e A = lim_xto a fraclog1+f(x)g(x)$$
$$implies log_e A = lim_xto a fraclog1+f(x)f(x) cdot fracf(x)g(x)$$
$$implies log_e A = lim_xto a fracf(x)g(x)$$
$$implies A=e^limlimits_xto a fracf(x)g(x)$$

Now my doubt here is... how did the first step become second step by taking log on both sides? How did the log go inside the limit? That is my confusion.

That is, $A=lim_xto a 1+f(x)^1over g(x)$ became $log_e A = lim_xto a fraclog1+f(x)g(x)$. How?

Hope you get what my doubt is. Thanks.

calculus limits

share|cite|improve this question

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

$endgroup$

add a comment | 

$begingroup$

There is a theorem which says "If $limlimits_xto a f(x) =limlimits_xto a g(x)=0$ such that $limlimits_xto a fracf(x)g(x)$ exists, then $limlimits_x to a 1+f(x)^frac1g(x) = e^limlimits_xto a fracf(x)g(x)$

The proof for this theorem goes as such:
$$textLet A=lim_xto a 1+f(x)^1over g(x).text Then, $$
$$log_e A = lim_xto a fraclog1+f(x)g(x)$$
$$implies log_e A = lim_xto a fraclog1+f(x)f(x) cdot fracf(x)g(x)$$
$$implies log_e A = lim_xto a fracf(x)g(x)$$
$$implies A=e^limlimits_xto a fracf(x)g(x)$$

Now my doubt here is... how did the first step become second step by taking log on both sides? How did the log go inside the limit? That is my confusion.

That is, $A=lim_xto a 1+f(x)^1over g(x)$ became $log_e A = lim_xto a fraclog1+f(x)g(x)$. How?

Hope you get what my doubt is. Thanks.

calculus limits

share|cite|improve this question

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

$endgroup$

share|cite|improve this question

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

share|cite|improve this question

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

edited May 13 at 15:54

Nick

318113

edited May 13 at 15:54

Nick

318113

asked May 13 at 14:07

rashrash

636216

asked May 13 at 14:07

rashrash

636216

2 Answers
2

active

oldest

votes

5

$begingroup$

Regarding your question about the first step: $ln$ is continuous, i.e. you can write the limit inside this function (given that it exists). The $g(x)$ in the denominator comes from the fact that $$ln(a^b)=bln(a),$$ where you can choose $a=1+f(x)$ and $b=1/g(x)$.

share|cite|improve this answer

answered May 13 at 14:13

st.mathst.math

1,453215

$endgroup$

add a comment | 

0

$begingroup$

The theorem is false: take the constant functions $f(x)=g(x)=1.$

share|cite|improve this answer

answered May 13 at 14:13

FredFred

49.7k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it false?
  $endgroup$
  – rash
  May 13 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @rash It is correct now, but it used to be false before you edited your question.
  $endgroup$
  – st.math
  May 13 at 14:24
  

  
  
  
  

add a comment | 

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5

$begingroup$

Regarding your question about the first step: $ln$ is continuous, i.e. you can write the limit inside this function (given that it exists). The $g(x)$ in the denominator comes from the fact that $$ln(a^b)=bln(a),$$ where you can choose $a=1+f(x)$ and $b=1/g(x)$.

share|cite|improve this answer

answered May 13 at 14:13

st.mathst.math

1,453215

$endgroup$

add a comment | 

5

$begingroup$

Regarding your question about the first step: $ln$ is continuous, i.e. you can write the limit inside this function (given that it exists). The $g(x)$ in the denominator comes from the fact that $$ln(a^b)=bln(a),$$ where you can choose $a=1+f(x)$ and $b=1/g(x)$.

share|cite|improve this answer

answered May 13 at 14:13

st.mathst.math

1,453215

$endgroup$

add a comment | 

$begingroup$

Regarding your question about the first step: $ln$ is continuous, i.e. you can write the limit inside this function (given that it exists). The $g(x)$ in the denominator comes from the fact that $$ln(a^b)=bln(a),$$ where you can choose $a=1+f(x)$ and $b=1/g(x)$.

share|cite|improve this answer

answered May 13 at 14:13

st.mathst.math

1,453215

$endgroup$

share|cite|improve this answer

answered May 13 at 14:13

st.mathst.math

1,453215

answered May 13 at 14:13

st.mathst.math

1,453215

answered May 13 at 14:13

st.mathst.math

1,453215

0

$begingroup$

The theorem is false: take the constant functions $f(x)=g(x)=1.$

share|cite|improve this answer

answered May 13 at 14:13

FredFred

49.7k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it false?
  $endgroup$
  – rash
  May 13 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @rash It is correct now, but it used to be false before you edited your question.
  $endgroup$
  – st.math
  May 13 at 14:24
  

  
  
  
  

add a comment | 

0

$begingroup$

The theorem is false: take the constant functions $f(x)=g(x)=1.$

share|cite|improve this answer

answered May 13 at 14:13

FredFred

49.7k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is it false?
  $endgroup$
  – rash
  May 13 at 14:19
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @rash It is correct now, but it used to be false before you edited your question.
  $endgroup$
  – st.math
  May 13 at 14:24
  

  
  
  
  

add a comment | 

$begingroup$

The theorem is false: take the constant functions $f(x)=g(x)=1.$

share|cite|improve this answer

answered May 13 at 14:13

FredFred

49.7k11849

$endgroup$

share|cite|improve this answer

answered May 13 at 14:13

FredFred

49.7k11849

answered May 13 at 14:13

FredFred

49.7k11849

answered May 13 at 14:13

FredFred

49.7k11849