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Evaluation of $lim_xtoinfty frac1xint_1^x fraclog(1+s)(1+log(s))sqrt1+s^2ds$ [closed]

Show that $lim_ntoinftyfracn+1n-2=1$inequality $ prod_n=2^inftyn^zeta(n)-1 <fracpi^2+66$Evaluate the limit $lim_nto infty sum _i=1^n (sqrtn+fracisqrtn)^-2$Solve the improper integral: $int_1^inftyfrac33e^-sqrtxsqrtx$Evaluation $lim_nto inftyfraclog^k nn^epsilon$Evaluation of the limit $lim_ntoinftysum_k=1^nfrac kk^2+n^2$How to evaluate $lim_ntoinfty(sqrtn^2+3n-n)$ and $lim_xto-infty(sqrt4x^2-2x+2x)$?Result of $lim_ntoinfty (1+ sqrtn+1 -sqrtn)^sqrtn$Limit $lim_xtoinftyxlogleft(fracx+7x+2right)$Find a vector which is perpendicular to both $ti+t²j−2k$ and $si+2j+s²k$.

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I really don't know how to start. I need an hint to get started.

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calculus limits analysis

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asked Mar 20 at 19:27

SimmetricoSimmetrico

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closed as off-topic by Paramanand Singh, Strants, clathratus, YiFan, zz20s Mar 22 at 2:44

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." – Paramanand Singh, Strants, clathratus, YiFan, zz20s

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

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  Show your integrand grows sub-linearly so the limit is zero.
  $endgroup$
  – Brevan Ellefsen
  Mar 20 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
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  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  42 mins ago
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I really don't know how to start. I need an hint to get started.

Thank you.

calculus limits analysis

share|cite|improve this question

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

$endgroup$

closed as off-topic by Paramanand Singh, Strants, clathratus, YiFan, zz20s Mar 22 at 2:44

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." – Paramanand Singh, Strants, clathratus, YiFan, zz20s

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Show your integrand grows sub-linearly so the limit is zero.
  $endgroup$
  – Brevan Ellefsen
  Mar 20 at 19:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  42 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I really don't know how to start. I need an hint to get started.

Thank you.

calculus limits analysis

share|cite|improve this question

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

$endgroup$

share|cite|improve this question

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

share|cite|improve this question

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

asked Mar 20 at 19:27

SimmetricoSimmetrico

93

2 Answers
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Hint: Note that $log(1+s)leq1+log s$ and $sleqsqrt1+s^2$, so you can bound the integrand from above by $frac1s$. You can integrate $frac1s$ and calculate the proposed limit.

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answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

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Using L'Hospital's Rule gives

$$lim_xto inftyfrac1x int_0^x fraclog(1+s)(1+log(s))sqrt1+s^2,ds=lim_xtoinftyfraclog(1+x)(1+log(x))sqrt1+x^2=0$$

Note that when the denominator approaches infinity, it is not required that the limit of the numerator approach infinity for LHR to be applicable provided all of the other conditions apply. In fact, the limit of the numerator need not even exist.

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answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @simmetrico Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:59
  
  

  
  
  
  

add a comment | 

0

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Hint: Note that $log(1+s)leq1+log s$ and $sleqsqrt1+s^2$, so you can bound the integrand from above by $frac1s$. You can integrate $frac1s$ and calculate the proposed limit.

share|cite|improve this answer

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

$endgroup$

add a comment | 

0

$begingroup$

Hint: Note that $log(1+s)leq1+log s$ and $sleqsqrt1+s^2$, so you can bound the integrand from above by $frac1s$. You can integrate $frac1s$ and calculate the proposed limit.

share|cite|improve this answer

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

$endgroup$

add a comment | 

$begingroup$

Hint: Note that $log(1+s)leq1+log s$ and $sleqsqrt1+s^2$, so you can bound the integrand from above by $frac1s$. You can integrate $frac1s$ and calculate the proposed limit.

share|cite|improve this answer

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

$endgroup$

share|cite|improve this answer

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

answered Mar 20 at 19:34

ClaytonClayton

19.6k33288

0

$begingroup$

Using L'Hospital's Rule gives

$$lim_xto inftyfrac1x int_0^x fraclog(1+s)(1+log(s))sqrt1+s^2,ds=lim_xtoinftyfraclog(1+x)(1+log(x))sqrt1+x^2=0$$

Note that when the denominator approaches infinity, it is not required that the limit of the numerator approach infinity for LHR to be applicable provided all of the other conditions apply. In fact, the limit of the numerator need not even exist.

share|cite|improve this answer

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @simmetrico Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:59
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Using L'Hospital's Rule gives

$$lim_xto inftyfrac1x int_0^x fraclog(1+s)(1+log(s))sqrt1+s^2,ds=lim_xtoinftyfraclog(1+x)(1+log(x))sqrt1+x^2=0$$

Note that when the denominator approaches infinity, it is not required that the limit of the numerator approach infinity for LHR to be applicable provided all of the other conditions apply. In fact, the limit of the numerator need not even exist.

share|cite|improve this answer

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @simmetrico Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote and accept an answer as you see fit.
  $endgroup$
  – Mark Viola
  Mar 25 at 17:59
  
  

  
  
  
  

add a comment | 

$begingroup$

Using L'Hospital's Rule gives

$$lim_xto inftyfrac1x int_0^x fraclog(1+s)(1+log(s))sqrt1+s^2,ds=lim_xtoinftyfraclog(1+x)(1+log(x))sqrt1+x^2=0$$

Note that when the denominator approaches infinity, it is not required that the limit of the numerator approach infinity for LHR to be applicable provided all of the other conditions apply. In fact, the limit of the numerator need not even exist.

share|cite|improve this answer

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

$endgroup$

share|cite|improve this answer

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176

answered Mar 20 at 20:13

Mark ViolaMark Viola

134k1278176