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Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$

0

2

$begingroup$

Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$

My trial

As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,

begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}

This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?

calculus algebra-precalculus limits

share|cite|improve this question

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/75130/…
  $endgroup$
  – Robert Z
  Mar 26 at 10:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z: Thanks a lot, I got that hint!
  $endgroup$
  – Omojola Micheal
  Mar 26 at 10:26
  

  
  
  
  

add a comment | 

0

2

$begingroup$

Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$

My trial

As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,

begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}

This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?

calculus algebra-precalculus limits

share|cite|improve this question

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See math.stackexchange.com/questions/75130/…
  $endgroup$
  – Robert Z
  Mar 26 at 10:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Robert Z: Thanks a lot, I got that hint!
  $endgroup$
  – Omojola Micheal
  Mar 26 at 10:26
  

  
  
  
  

add a comment | 

$begingroup$

Evaluate $limlimits_{xto 0^{+}}(ln(x)-ln(sin x))$

My trial

As $xto 0^{+},;ln(x)to infty$ and $ln(sin x)to infty.$ So,

begin{align}limlimits_{xto 0^{+}}(ln(x)-ln(sin x))=limlimits_{xto 0^{+}}lnleft(frac{x}{sin x}right)end{align}

This should result to $infty$ but I may be wrong. If I am wrong, how do I apply L'Hopital's rule to this?

calculus algebra-precalculus limits

share|cite|improve this question

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

$endgroup$

share|cite|improve this question

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

share|cite|improve this question

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

asked Mar 26 at 10:21

Omojola MichealOmojola Micheal

2,082424

2 Answers
2

active

oldest

votes

5

$begingroup$

$ frac{x}{sin x} to 1$ as $x to 0$,

hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$

share|cite|improve this answer

answered Mar 26 at 10:24

FredFred

48.6k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thanks Fred!
  $endgroup$
  – Omojola Micheal
  Mar 26 at 10:25
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint:

1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


2. 
   $ln$ is continuous.


3. 
   $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.

share|cite|improve this answer

answered Mar 26 at 10:26

5xum5xum

92.9k395162

$endgroup$

add a comment | 

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5

$begingroup$

$ frac{x}{sin x} to 1$ as $x to 0$,

hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$

share|cite|improve this answer

answered Mar 26 at 10:24

FredFred

48.6k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thanks Fred!
  $endgroup$
  – Omojola Micheal
  Mar 26 at 10:25
  

  
  
  
  

add a comment | 

5

$begingroup$

$ frac{x}{sin x} to 1$ as $x to 0$,

hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$

share|cite|improve this answer

answered Mar 26 at 10:24

FredFred

48.6k11849

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thanks Fred!
  $endgroup$
  – Omojola Micheal
  Mar 26 at 10:25
  

  
  
  
  

add a comment | 

$begingroup$

$ frac{x}{sin x} to 1$ as $x to 0$,

hence $ ln (frac{x}{sin x}) to ln 1=0$ as $x to 0.$

share|cite|improve this answer

answered Mar 26 at 10:24

FredFred

48.6k11849

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:24

FredFred

48.6k11849

answered Mar 26 at 10:24

FredFred

48.6k11849

answered Mar 26 at 10:24

FredFred

48.6k11849

2

$begingroup$

Hint:

1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


2. 
   $ln$ is continuous.


3. 
   $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.

share|cite|improve this answer

answered Mar 26 at 10:26

5xum5xum

92.9k395162

$endgroup$

add a comment | 

2

$begingroup$

Hint:

1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


2. 
   $ln$ is continuous.


3. 
   $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.

share|cite|improve this answer

answered Mar 26 at 10:26

5xum5xum

92.9k395162

$endgroup$

add a comment | 

$begingroup$

Hint:

1. If $f$ is a continuous function and $lim_{xto a} g(x) = L$, then $lim_{xto a} f(g(x)) = f(L)$.


2. 
   $ln$ is continuous.


3. 
   $lim_{xto 0}frac{x}{sin x}$ is simple to calculate.

share|cite|improve this answer

answered Mar 26 at 10:26

5xum5xum

92.9k395162

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:26

5xum5xum

92.9k395162

answered Mar 26 at 10:26

5xum5xum

92.9k395162

answered Mar 26 at 10:26

5xum5xum

92.9k395162