How to find trigonometric limit $ lim_{xto π} frac{sin x}{x^2 - π^2} $ [closed]

-1

$begingroup$

how to solve this problem? (without using l'Hopital rule)
$$ lim_{xto π} frac{sin x}{x^2 - π^2} $$
I have no idea how to transform that so that I have no expression $frac{0}{0}$

I suspect that calculating limit $lim_{xto π^{-}}$ or $lim_{xto π^+}$ does not works there.


Thanks for helping

trigonometry limits-without-lhopital

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asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

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closed as off-topic by Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy Jan 3 at 23:17

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." – Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy

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

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
  $endgroup$
  – Jack D'Aurizio
  Jan 1 at 22:24
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

how to solve this problem? (without using l'Hopital rule)
$$ lim_{xto π} frac{sin x}{x^2 - π^2} $$
I have no idea how to transform that so that I have no expression $frac{0}{0}$

I suspect that calculating limit $lim_{xto π^{-}}$ or $lim_{xto π^+}$ does not works there.


Thanks for helping

trigonometry limits-without-lhopital

share|cite|improve this question

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

$endgroup$

closed as off-topic by Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy Jan 3 at 23:17

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." – Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy

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

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
  $endgroup$
  – Jack D'Aurizio
  Jan 1 at 22:24
  
  
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

how to solve this problem? (without using l'Hopital rule)
$$ lim_{xto π} frac{sin x}{x^2 - π^2} $$
I have no idea how to transform that so that I have no expression $frac{0}{0}$

I suspect that calculating limit $lim_{xto π^{-}}$ or $lim_{xto π^+}$ does not works there.


Thanks for helping

trigonometry limits-without-lhopital

share|cite|improve this question

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

$endgroup$

how to solve this problem? (without using l'Hopital rule)
$$ lim_{xto π} frac{sin x}{x^2 - π^2} $$
I have no idea how to transform that so that I have no expression $frac{0}{0}$

I suspect that calculating limit $lim_{xto π^{-}}$ or $lim_{xto π^+}$ does not works there.


Thanks for helping

trigonometry limits-without-lhopital

trigonometry limits-without-lhopital

share|cite|improve this question

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

share|cite|improve this question

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

share|cite|improve this question

share|cite|improve this question

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

asked Jan 1 at 22:23

VirtualUserVirtualUser

59412

59412

closed as off-topic by Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy Jan 3 at 23:17

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." – Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy

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

closed as off-topic by Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy Jan 3 at 23:17

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." – Nosrati, user91500, mrtaurho, Lee David Chung Lin, amWhy

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

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
  $endgroup$
  – Jack D'Aurizio
  Jan 1 at 22:24
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  9
  

  
  
  
  
  
  

  
  $begingroup$
  Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
  $endgroup$
  – Jack D'Aurizio
  Jan 1 at 22:24
  
  
  

  
  
  
  

9

9

$begingroup$
Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
$endgroup$
– Jack D'Aurizio
Jan 1 at 22:24

$begingroup$
Let $x=pi+y$ and wonder about $$lim_{yto 0}frac{-color{blue}{sin y}}{color{blue}{y}color{red}{(2pi+y)}}.$$
$endgroup$
– Jack D'Aurizio
Jan 1 at 22:24

add a comment | 

2 Answers
2

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11

$begingroup$

Rewrite as
$$
lim_{xto pi}frac{sin x}{x^2-pi^2}=lim_{xto pi}frac{sin x-sin pi}{x-pi}times
lim_{xto pi}frac{1}{x+pi}
$$
The first limit is (by definition) the derivative of $sin $ at $x=pi$. The second limit is easy to compute.

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edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

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0

$begingroup$

Hint: $$lim_{xtopi}frac{sin x}{x^2-pi^2}=left(lim_{xtopi}frac1{x+pi}right)left(lim_{xtopi}frac{sin x}{x-pi}right).$$

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answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

11

$begingroup$

Rewrite as
$$
lim_{xto pi}frac{sin x}{x^2-pi^2}=lim_{xto pi}frac{sin x-sin pi}{x-pi}times
lim_{xto pi}frac{1}{x+pi}
$$
The first limit is (by definition) the derivative of $sin $ at $x=pi$. The second limit is easy to compute.

share|cite|improve this answer

edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

$endgroup$

add a comment | 

11

$begingroup$

Rewrite as
$$
lim_{xto pi}frac{sin x}{x^2-pi^2}=lim_{xto pi}frac{sin x-sin pi}{x-pi}times
lim_{xto pi}frac{1}{x+pi}
$$
The first limit is (by definition) the derivative of $sin $ at $x=pi$. The second limit is easy to compute.

share|cite|improve this answer

edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

$endgroup$

add a comment | 

11

11

11

$begingroup$

Rewrite as
$$
lim_{xto pi}frac{sin x}{x^2-pi^2}=lim_{xto pi}frac{sin x-sin pi}{x-pi}times
lim_{xto pi}frac{1}{x+pi}
$$
The first limit is (by definition) the derivative of $sin $ at $x=pi$. The second limit is easy to compute.

share|cite|improve this answer

edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

$endgroup$

Rewrite as
$$
lim_{xto pi}frac{sin x}{x^2-pi^2}=lim_{xto pi}frac{sin x-sin pi}{x-pi}times
lim_{xto pi}frac{1}{x+pi}
$$
The first limit is (by definition) the derivative of $sin $ at $x=pi$. The second limit is easy to compute.

share|cite|improve this answer

edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

share|cite|improve this answer

share|cite|improve this answer

edited Jan 1 at 22:33

edited Jan 1 at 22:33

edited Jan 1 at 22:33

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

answered Jan 1 at 22:25

Foobaz JohnFoobaz John

21.6k41351

21.6k41351

add a comment | 

add a comment | 

0

$begingroup$

Hint: $$lim_{xtopi}frac{sin x}{x^2-pi^2}=left(lim_{xtopi}frac1{x+pi}right)left(lim_{xtopi}frac{sin x}{x-pi}right).$$

share|cite|improve this answer

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

$endgroup$

add a comment | 

0

$begingroup$

Hint: $$lim_{xtopi}frac{sin x}{x^2-pi^2}=left(lim_{xtopi}frac1{x+pi}right)left(lim_{xtopi}frac{sin x}{x-pi}right).$$

share|cite|improve this answer

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint: $$lim_{xtopi}frac{sin x}{x^2-pi^2}=left(lim_{xtopi}frac1{x+pi}right)left(lim_{xtopi}frac{sin x}{x-pi}right).$$

share|cite|improve this answer

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

$endgroup$

Hint: $$lim_{xtopi}frac{sin x}{x^2-pi^2}=left(lim_{xtopi}frac1{x+pi}right)left(lim_{xtopi}frac{sin x}{x-pi}right).$$

share|cite|improve this answer

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

share|cite|improve this answer

share|cite|improve this answer

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

answered Jan 1 at 22:25

José Carlos SantosJosé Carlos Santos

153k22123225

153k22123225

add a comment | 

add a comment | 

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