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Showing a limit approaches e: base of natural log

About $lim left(1+frac xnright)^n$Solving a tough limitUsing L'Hospital's Rule to evaluate limit to infinityLimit of Lambert $W$ Product Log is the Natural Log?I want to calculate the limit of: $lim_x to 0 left(frac2^x+8^x2 right)^frac1x $I would like to calculate the following limit: $lim_ n to infty left( ncdot sinfrac1n right)^n^2$Proof of a limit for a recursively-defined sequenceLimit, Sum, and Natural LogLimit as a definite integral (Riemann Sum)Evaluate the limit of $frac1log(x)$

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3

1

$begingroup$

I am in an intro to Analysis class, and I want to show that

$$lim_nto infty left( 1+frac12n right)^n = sqrte$$

I already have a result that

$$lim_nto infty left( 1+frac1n right)^n = e$$

Which was not proven rigorously but was given to us. We basically just proved that the limit was bounded between 2 and 3, and then they gave us the real limit $e$. I am not sure if I need to use that result or not.

I have also found in the text a property that says a suite of real numbers converges to $sqrta$ if $x_1>0$ and $x_n=frac12left(x_n-1+fracax_n-1right),nge2$

I am having trouble representing the first expression in this way. Any direction would be greatly appreciated.

real-analysis limits eulers-number

share|cite|improve this question

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note: A limit doesn't approach anything. It's not moving around and doing a little dance. It either exists or it doesn't, and if it does, it's just a number.
  $endgroup$
  – zhw.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of About $lim left(1+frac xnright)^n$
  $endgroup$
  – JoeTaxpayer
  1 hour ago
  

  
  
  
  

add a comment
 | 

3

1

$begingroup$

I am in an intro to Analysis class, and I want to show that

$$lim_nto infty left( 1+frac12n right)^n = sqrte$$

I already have a result that

$$lim_nto infty left( 1+frac1n right)^n = e$$

Which was not proven rigorously but was given to us. We basically just proved that the limit was bounded between 2 and 3, and then they gave us the real limit $e$. I am not sure if I need to use that result or not.

I have also found in the text a property that says a suite of real numbers converges to $sqrta$ if $x_1>0$ and $x_n=frac12left(x_n-1+fracax_n-1right),nge2$

I am having trouble representing the first expression in this way. Any direction would be greatly appreciated.

real-analysis limits eulers-number

share|cite|improve this question

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note: A limit doesn't approach anything. It's not moving around and doing a little dance. It either exists or it doesn't, and if it does, it's just a number.
  $endgroup$
  – zhw.
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of About $lim left(1+frac xnright)^n$
  $endgroup$
  – JoeTaxpayer
  1 hour ago
  

  
  
  
  

add a comment
 | 

$begingroup$

I am in an intro to Analysis class, and I want to show that

$$lim_nto infty left( 1+frac12n right)^n = sqrte$$

I already have a result that

$$lim_nto infty left( 1+frac1n right)^n = e$$

Which was not proven rigorously but was given to us. We basically just proved that the limit was bounded between 2 and 3, and then they gave us the real limit $e$. I am not sure if I need to use that result or not.

I have also found in the text a property that says a suite of real numbers converges to $sqrta$ if $x_1>0$ and $x_n=frac12left(x_n-1+fracax_n-1right),nge2$

I am having trouble representing the first expression in this way. Any direction would be greatly appreciated.

real-analysis limits eulers-number

share|cite|improve this question

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

$endgroup$

share|cite|improve this question

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

share|cite|improve this question

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

edited 10 hours ago

José Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

asked 10 hours ago

jeffery_the_windjeffery_the_wind

16799 bronze badges

3 Answers
3

active

oldest

votes

6

$begingroup$

We have that

$$left( 1+frac12nright)^n=left[left( 1+frac12nright)^2nright]^frac12$$

then let $k=2n to infty$.

share|cite|improve this answer

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is very easy :-). +1 for all the users.
  $endgroup$
  – Sebastiano
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thanks for the quick explanation, but I am in an intro class and we are expected to show every part of the logic. I see what you did here but I am missing that part that we are allowed to assume that the limit of a product of a suite approaches the product of the limits of the component suites.
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK I guess I found it here: if $lim_nto infty x_n = x$ and $lim_nto infty y_n = y$ then $lim_nto infty x_ny_n = xy$
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jeffery_the_wind Yes of course! Refer to these notes.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  

add a comment
 | 

6

$begingroup$

Hint:$$left(1+frac12nright)^n=sqrtleft(1+frac12nright)^2n.$$

share|cite|improve this answer

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

$$
left(1+frac 12nright)^2nlongrightarrow e.
$$
Now, by the continuity of the square root,
$$
left(1+frac 12nright)^n = left[left(1+frac 12nright)^2nright]^1/2longrightarrow sqrt e.
$$

share|cite|improve this answer

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

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add a comment
 | 

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6

$begingroup$

We have that

$$left( 1+frac12nright)^n=left[left( 1+frac12nright)^2nright]^frac12$$

then let $k=2n to infty$.

share|cite|improve this answer

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is very easy :-). +1 for all the users.
  $endgroup$
  – Sebastiano
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thanks for the quick explanation, but I am in an intro class and we are expected to show every part of the logic. I see what you did here but I am missing that part that we are allowed to assume that the limit of a product of a suite approaches the product of the limits of the component suites.
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK I guess I found it here: if $lim_nto infty x_n = x$ and $lim_nto infty y_n = y$ then $lim_nto infty x_ny_n = xy$
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jeffery_the_wind Yes of course! Refer to these notes.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  

add a comment
 | 

6

$begingroup$

We have that

$$left( 1+frac12nright)^n=left[left( 1+frac12nright)^2nright]^frac12$$

then let $k=2n to infty$.

share|cite|improve this answer

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is very easy :-). +1 for all the users.
  $endgroup$
  – Sebastiano
  10 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes thanks for the quick explanation, but I am in an intro class and we are expected to show every part of the logic. I see what you did here but I am missing that part that we are allowed to assume that the limit of a product of a suite approaches the product of the limits of the component suites.
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK I guess I found it here: if $lim_nto infty x_n = x$ and $lim_nto infty y_n = y$ then $lim_nto infty x_ny_n = xy$
  $endgroup$
  – jeffery_the_wind
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jeffery_the_wind Yes of course! Refer to these notes.
  $endgroup$
  – gimusi
  10 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

We have that

$$left( 1+frac12nright)^n=left[left( 1+frac12nright)^2nright]^frac12$$

then let $k=2n to infty$.

share|cite|improve this answer

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

answered 10 hours ago

gimusigimusi

96.8k99 gold badges4646 silver badges9595 bronze badges

6

$begingroup$

Hint:$$left(1+frac12nright)^n=sqrtleft(1+frac12nright)^2n.$$

share|cite|improve this answer

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

Hint:$$left(1+frac12nright)^n=sqrtleft(1+frac12nright)^2n.$$

share|cite|improve this answer

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Hint:$$left(1+frac12nright)^n=sqrtleft(1+frac12nright)^2n.$$

share|cite|improve this answer

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

answered 10 hours ago

José Carlos SantosJosé Carlos Santos

219k2626 gold badges171171 silver badges294294 bronze badges

5

$begingroup$

$$
left(1+frac 12nright)^2nlongrightarrow e.
$$
Now, by the continuity of the square root,
$$
left(1+frac 12nright)^n = left[left(1+frac 12nright)^2nright]^1/2longrightarrow sqrt e.
$$

share|cite|improve this answer

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

$$
left(1+frac 12nright)^2nlongrightarrow e.
$$
Now, by the continuity of the square root,
$$
left(1+frac 12nright)^n = left[left(1+frac 12nright)^2nright]^1/2longrightarrow sqrt e.
$$

share|cite|improve this answer

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

$$
left(1+frac 12nright)^2nlongrightarrow e.
$$
Now, by the continuity of the square root,
$$
left(1+frac 12nright)^n = left[left(1+frac 12nright)^2nright]^1/2longrightarrow sqrt e.
$$

share|cite|improve this answer

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

$endgroup$

share|cite|improve this answer

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges

answered 10 hours ago

amsmathamsmath

4,83277 silver badges2222 bronze badges