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Using L'Hospital's Rule on $cos (2x)^frac 1 x^2.$ [closed]

-3

1

$begingroup$

How might one evaluate the following limit using L'Hopital's Rule:$$lim_xrightarrow0cos(2x)^frac1x^2$$

limits analysis

share|cite|improve this question

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

$endgroup$

closed as off-topic by RRL, Eevee Trainer, mrtaurho, Cesareo, Javi Apr 1 at 10:33

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." – RRL, Eevee Trainer, mrtaurho, Cesareo, Javi

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Kezer
  Mar 30 at 9:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Arrived at $e^-2$ using $e^lim_xrightarrow0 fracln(cos(2x))x^2$
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:57
  
  

  
  
  
  

add a comment | 

-3

1

$begingroup$

How might one evaluate the following limit using L'Hopital's Rule:$$lim_xrightarrow0cos(2x)^frac1x^2$$

limits analysis

share|cite|improve this question

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

$endgroup$

closed as off-topic by RRL, Eevee Trainer, mrtaurho, Cesareo, Javi Apr 1 at 10:33

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." – RRL, Eevee Trainer, mrtaurho, Cesareo, Javi

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Kezer
  Mar 30 at 9:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Arrived at $e^-2$ using $e^lim_xrightarrow0 fracln(cos(2x))x^2$
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:57
  
  

  
  
  
  

add a comment | 

$begingroup$

How might one evaluate the following limit using L'Hopital's Rule:$$lim_xrightarrow0cos(2x)^frac1x^2$$

limits analysis

share|cite|improve this question

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

$endgroup$

share|cite|improve this question

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

share|cite|improve this question

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

edited Mar 30 at 10:19

Bernard

124k741117

edited Mar 30 at 10:19

Bernard

124k741117

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

asked Mar 30 at 9:29

Ali LodhiAli Lodhi

383

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint: Write $$e^lim_xto 0fracln(cos(2x)x^2$$

share|cite|improve this answer

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, thankyou.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Want you to know the result?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I want to know the result.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$e^-2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, arrived at that. Thank you.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:53
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:

$$y=cos(2x)^1/x^2Rightarrow ln y=4fraclncos(2x)(2x)^2\
1-frac1zleqln zleq z-1$$

share|cite|improve this answer

answered Mar 30 at 9:58

JamJam

5,01921432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proof of bounds: proofwiki.org/wiki/Bounds_of_Natural_Logarithm
  $endgroup$
  – Jam
  Mar 30 at 9:59
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Write $$e^lim_xto 0fracln(cos(2x)x^2$$

share|cite|improve this answer

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, thankyou.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Want you to know the result?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I want to know the result.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$e^-2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, arrived at that. Thank you.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:53
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint: Write $$e^lim_xto 0fracln(cos(2x)x^2$$

share|cite|improve this answer

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, thankyou.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Want you to know the result?
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I want to know the result.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is $$e^-2$$
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 30 at 9:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, arrived at that. Thank you.
  $endgroup$
  – Ali Lodhi
  Mar 30 at 9:53
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Write $$e^lim_xto 0fracln(cos(2x)x^2$$

share|cite|improve this answer

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

answered Mar 30 at 9:31

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

78.8k42867

0

$begingroup$

Hint:

$$y=cos(2x)^1/x^2Rightarrow ln y=4fraclncos(2x)(2x)^2\
1-frac1zleqln zleq z-1$$

share|cite|improve this answer

answered Mar 30 at 9:58

JamJam

5,01921432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proof of bounds: proofwiki.org/wiki/Bounds_of_Natural_Logarithm
  $endgroup$
  – Jam
  Mar 30 at 9:59
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:

$$y=cos(2x)^1/x^2Rightarrow ln y=4fraclncos(2x)(2x)^2\
1-frac1zleqln zleq z-1$$

share|cite|improve this answer

answered Mar 30 at 9:58

JamJam

5,01921432

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Proof of bounds: proofwiki.org/wiki/Bounds_of_Natural_Logarithm
  $endgroup$
  – Jam
  Mar 30 at 9:59
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

$$y=cos(2x)^1/x^2Rightarrow ln y=4fraclncos(2x)(2x)^2\
1-frac1zleqln zleq z-1$$

share|cite|improve this answer

answered Mar 30 at 9:58

JamJam

5,01921432

$endgroup$

share|cite|improve this answer

answered Mar 30 at 9:58

JamJam

5,01921432

answered Mar 30 at 9:58

JamJam

5,01921432

answered Mar 30 at 9:58

JamJam

5,01921432