Real Limit Problem Found in a Scholarship Exam for a Masters C0urse [closed]

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Evaluate: $lim_xrightarrow0(cos x)^frac1sin^2x$.

I found this problem in one of the previous years papers of a Scholarship exam I'll be giving soon. Any hints are welcome!

real-analysis limits

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asked Jul 17 at 15:31

Samuel Davidson

12

closed as off-topic by Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus Jul 18 at 1:41

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You'll be giving, or you'll be taking?
  – zhw.
  Jul 17 at 20:50
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

Evaluate: $lim_xrightarrow0(cos x)^frac1sin^2x$.

I found this problem in one of the previous years papers of a Scholarship exam I'll be giving soon. Any hints are welcome!

real-analysis limits

share|cite|improve this question

asked Jul 17 at 15:31

Samuel Davidson

12

closed as off-topic by Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus Jul 18 at 1:41

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You'll be giving, or you'll be taking?
  – zhw.
  Jul 17 at 20:50
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Evaluate: $lim_xrightarrow0(cos x)^frac1sin^2x$.

I found this problem in one of the previous years papers of a Scholarship exam I'll be giving soon. Any hints are welcome!

real-analysis limits

share|cite|improve this question

asked Jul 17 at 15:31

Samuel Davidson

12

Evaluate: $lim_xrightarrow0(cos x)^frac1sin^2x$.

I found this problem in one of the previous years papers of a Scholarship exam I'll be giving soon. Any hints are welcome!

real-analysis limits

share|cite|improve this question

asked Jul 17 at 15:31

Samuel Davidson

12

share|cite|improve this question

share|cite|improve this question

asked Jul 17 at 15:31

Samuel Davidson

12

asked Jul 17 at 15:31

Samuel Davidson

12

asked Jul 17 at 15:31

Samuel Davidson

12

12

closed as off-topic by Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus Jul 18 at 1:41

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus Jul 18 at 1:41

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, steven gregory, amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You'll be giving, or you'll be taking?
  – zhw.
  Jul 17 at 20:50
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You'll be giving, or you'll be taking?
  – zhw.
  Jul 17 at 20:50
  

  
  
  
  

You'll be giving, or you'll be taking?
– zhw.
Jul 17 at 20:50

You'll be giving, or you'll be taking?
– zhw.
Jul 17 at 20:50

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote

Hint:

Introduce substitution $$sin x=frac1u$$

You will get a limit which looks like the famous definition of $e$.

The result is $frac1sqrt e$. This is all fairly simple, it's better to try yourself and learn something along the way.

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

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up vote
1
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When $1>cos x >0$ we have $ln ((cos x)^(1/sin^2 x))=$ $frac 1sin^2 xln cos x=$ $frac 1u^2ln sqrt 1-u^2$ where $u=sin x.$

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answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

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up vote
1
down vote

$$cos(x)^frac1sin^2(x)
= expleft(fraclncos(x)sin^2(x)right)
= expleft(fraclncos(x)tfrac12[1-cos(2x)]right)$$

Dealing with the fractional expression, recall the approximations of $cos$ and $ln$ centered around $0$ and $1$, respectively:
$$cos(x) = 1 - x^2/2 +O(x^4)$$

$$ln (1-x) = -x+O(x^2)$$
then for the numerator we get
beginalign
lncos(x)
&= ln(1-x^2/2+O(x^4))\
&= -(x^2/2-O(x^4)) + Obig((x^2/2-O(x^4))^2big)\
&= -tfrac12x^2 + O(x^4)
endalign
Hence, by continuity of exponentiation,
beginalign
lim_xto 0 cos(x)^frac1sin^2(x)
&= expleft( lim_xto 0 frac2lncos(x)1-cos(2x) right)\
&=expleft( lim_xto 0 frac2big(-tfrac12 x^2 + O(x^4)big)1 - big(1-(2x)^2/2+O((2x)^4)big) right)\
&=expleft( lim_xto 0 frac-x^2 + O(x^4)2x^2+O(x^4) right)
=expleft( -tfrac12 right)\
endalign

share|cite|improve this answer

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
  – Oldboy
  Jul 17 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Oldboy Missed a sign, cheers.
  – adfriedman
  Jul 17 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for being original, this is by far the most complicated way to solve this limit :)
  – Oldboy
  Jul 17 at 20:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was about 4 steps on paper. I answer in excruciating detail, just in case.
  – adfriedman
  Jul 17 at 21:32
  

  
  
  
  

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Hint:

Introduce substitution $$sin x=frac1u$$

You will get a limit which looks like the famous definition of $e$.

The result is $frac1sqrt e$. This is all fairly simple, it's better to try yourself and learn something along the way.

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

add a comment | 

up vote
3
down vote

Hint:

Introduce substitution $$sin x=frac1u$$

You will get a limit which looks like the famous definition of $e$.

The result is $frac1sqrt e$. This is all fairly simple, it's better to try yourself and learn something along the way.

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

add a comment | 

up vote
3
down vote

up vote
3
down vote

Hint:

Introduce substitution $$sin x=frac1u$$

You will get a limit which looks like the famous definition of $e$.

The result is $frac1sqrt e$. This is all fairly simple, it's better to try yourself and learn something along the way.

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

Hint:

Introduce substitution $$sin x=frac1u$$

You will get a limit which looks like the famous definition of $e$.

The result is $frac1sqrt e$. This is all fairly simple, it's better to try yourself and learn something along the way.

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 15:50

Oldboy

2,6101316

answered Jul 17 at 15:50

Oldboy

2,6101316

answered Jul 17 at 15:50

Oldboy

2,6101316

2,6101316

add a comment | 

add a comment | 

up vote
1
down vote

When $1>cos x >0$ we have $ln ((cos x)^(1/sin^2 x))=$ $frac 1sin^2 xln cos x=$ $frac 1u^2ln sqrt 1-u^2$ where $u=sin x.$

share|cite|improve this answer

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

add a comment | 

up vote
1
down vote

When $1>cos x >0$ we have $ln ((cos x)^(1/sin^2 x))=$ $frac 1sin^2 xln cos x=$ $frac 1u^2ln sqrt 1-u^2$ where $u=sin x.$

share|cite|improve this answer

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

add a comment | 

up vote
1
down vote

up vote
1
down vote

When $1>cos x >0$ we have $ln ((cos x)^(1/sin^2 x))=$ $frac 1sin^2 xln cos x=$ $frac 1u^2ln sqrt 1-u^2$ where $u=sin x.$

share|cite|improve this answer

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

When $1>cos x >0$ we have $ln ((cos x)^(1/sin^2 x))=$ $frac 1sin^2 xln cos x=$ $frac 1u^2ln sqrt 1-u^2$ where $u=sin x.$

share|cite|improve this answer

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

answered Jul 17 at 16:33

DanielWainfleet

31.7k31644

31.7k31644

add a comment | 

add a comment | 

up vote
1
down vote

$$cos(x)^frac1sin^2(x)
= expleft(fraclncos(x)sin^2(x)right)
= expleft(fraclncos(x)tfrac12[1-cos(2x)]right)$$

Dealing with the fractional expression, recall the approximations of $cos$ and $ln$ centered around $0$ and $1$, respectively:
$$cos(x) = 1 - x^2/2 +O(x^4)$$

$$ln (1-x) = -x+O(x^2)$$
then for the numerator we get
beginalign
lncos(x)
&= ln(1-x^2/2+O(x^4))\
&= -(x^2/2-O(x^4)) + Obig((x^2/2-O(x^4))^2big)\
&= -tfrac12x^2 + O(x^4)
endalign
Hence, by continuity of exponentiation,
beginalign
lim_xto 0 cos(x)^frac1sin^2(x)
&= expleft( lim_xto 0 frac2lncos(x)1-cos(2x) right)\
&=expleft( lim_xto 0 frac2big(-tfrac12 x^2 + O(x^4)big)1 - big(1-(2x)^2/2+O((2x)^4)big) right)\
&=expleft( lim_xto 0 frac-x^2 + O(x^4)2x^2+O(x^4) right)
=expleft( -tfrac12 right)\
endalign

share|cite|improve this answer

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
  – Oldboy
  Jul 17 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Oldboy Missed a sign, cheers.
  – adfriedman
  Jul 17 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for being original, this is by far the most complicated way to solve this limit :)
  – Oldboy
  Jul 17 at 20:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was about 4 steps on paper. I answer in excruciating detail, just in case.
  – adfriedman
  Jul 17 at 21:32
  

  
  
  
  

add a comment | 

up vote
1
down vote

$$cos(x)^frac1sin^2(x)
= expleft(fraclncos(x)sin^2(x)right)
= expleft(fraclncos(x)tfrac12[1-cos(2x)]right)$$

Dealing with the fractional expression, recall the approximations of $cos$ and $ln$ centered around $0$ and $1$, respectively:
$$cos(x) = 1 - x^2/2 +O(x^4)$$

$$ln (1-x) = -x+O(x^2)$$
then for the numerator we get
beginalign
lncos(x)
&= ln(1-x^2/2+O(x^4))\
&= -(x^2/2-O(x^4)) + Obig((x^2/2-O(x^4))^2big)\
&= -tfrac12x^2 + O(x^4)
endalign
Hence, by continuity of exponentiation,
beginalign
lim_xto 0 cos(x)^frac1sin^2(x)
&= expleft( lim_xto 0 frac2lncos(x)1-cos(2x) right)\
&=expleft( lim_xto 0 frac2big(-tfrac12 x^2 + O(x^4)big)1 - big(1-(2x)^2/2+O((2x)^4)big) right)\
&=expleft( lim_xto 0 frac-x^2 + O(x^4)2x^2+O(x^4) right)
=expleft( -tfrac12 right)\
endalign

share|cite|improve this answer

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
  – Oldboy
  Jul 17 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Oldboy Missed a sign, cheers.
  – adfriedman
  Jul 17 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for being original, this is by far the most complicated way to solve this limit :)
  – Oldboy
  Jul 17 at 20:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was about 4 steps on paper. I answer in excruciating detail, just in case.
  – adfriedman
  Jul 17 at 21:32
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

$$cos(x)^frac1sin^2(x)
= expleft(fraclncos(x)sin^2(x)right)
= expleft(fraclncos(x)tfrac12[1-cos(2x)]right)$$

Dealing with the fractional expression, recall the approximations of $cos$ and $ln$ centered around $0$ and $1$, respectively:
$$cos(x) = 1 - x^2/2 +O(x^4)$$

$$ln (1-x) = -x+O(x^2)$$
then for the numerator we get
beginalign
lncos(x)
&= ln(1-x^2/2+O(x^4))\
&= -(x^2/2-O(x^4)) + Obig((x^2/2-O(x^4))^2big)\
&= -tfrac12x^2 + O(x^4)
endalign
Hence, by continuity of exponentiation,
beginalign
lim_xto 0 cos(x)^frac1sin^2(x)
&= expleft( lim_xto 0 frac2lncos(x)1-cos(2x) right)\
&=expleft( lim_xto 0 frac2big(-tfrac12 x^2 + O(x^4)big)1 - big(1-(2x)^2/2+O((2x)^4)big) right)\
&=expleft( lim_xto 0 frac-x^2 + O(x^4)2x^2+O(x^4) right)
=expleft( -tfrac12 right)\
endalign

share|cite|improve this answer

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

$$cos(x)^frac1sin^2(x)
= expleft(fraclncos(x)sin^2(x)right)
= expleft(fraclncos(x)tfrac12[1-cos(2x)]right)$$

Dealing with the fractional expression, recall the approximations of $cos$ and $ln$ centered around $0$ and $1$, respectively:
$$cos(x) = 1 - x^2/2 +O(x^4)$$

$$ln (1-x) = -x+O(x^2)$$
then for the numerator we get
beginalign
lncos(x)
&= ln(1-x^2/2+O(x^4))\
&= -(x^2/2-O(x^4)) + Obig((x^2/2-O(x^4))^2big)\
&= -tfrac12x^2 + O(x^4)
endalign
Hence, by continuity of exponentiation,
beginalign
lim_xto 0 cos(x)^frac1sin^2(x)
&= expleft( lim_xto 0 frac2lncos(x)1-cos(2x) right)\
&=expleft( lim_xto 0 frac2big(-tfrac12 x^2 + O(x^4)big)1 - big(1-(2x)^2/2+O((2x)^4)big) right)\
&=expleft( lim_xto 0 frac-x^2 + O(x^4)2x^2+O(x^4) right)
=expleft( -tfrac12 right)\
endalign

share|cite|improve this answer

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 20:47

edited Jul 17 at 20:47

edited Jul 17 at 20:47

answered Jul 17 at 19:36

adfriedman

2,947169

answered Jul 17 at 19:36

adfriedman

2,947169

answered Jul 17 at 19:36

adfriedman

2,947169

2,947169

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
  – Oldboy
  Jul 17 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Oldboy Missed a sign, cheers.
  – adfriedman
  Jul 17 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for being original, this is by far the most complicated way to solve this limit :)
  – Oldboy
  Jul 17 at 20:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was about 4 steps on paper. I answer in excruciating detail, just in case.
  – adfriedman
  Jul 17 at 21:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
  – Oldboy
  Jul 17 at 20:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Oldboy Missed a sign, cheers.
  – adfriedman
  Jul 17 at 20:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for being original, this is by far the most complicated way to solve this limit :)
  – Oldboy
  Jul 17 at 20:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  It was about 4 steps on paper. I answer in excruciating detail, just in case.
  – adfriedman
  Jul 17 at 21:32
  

  
  
  
  

I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
– Oldboy
Jul 17 at 20:38

I will not downvote your answer but this is not the correct, the result should be $e^-1/2$.
– Oldboy
Jul 17 at 20:38

@Oldboy Missed a sign, cheers.
– adfriedman
Jul 17 at 20:47

@Oldboy Missed a sign, cheers.
– adfriedman
Jul 17 at 20:47

+1 for being original, this is by far the most complicated way to solve this limit :)
– Oldboy
Jul 17 at 20:56

+1 for being original, this is by far the most complicated way to solve this limit :)
– Oldboy
Jul 17 at 20:56

It was about 4 steps on paper. I answer in excruciating detail, just in case.
– adfriedman
Jul 17 at 21:32

It was about 4 steps on paper. I answer in excruciating detail, just in case.
– adfriedman
Jul 17 at 21:32

add a comment |