Limit involving integrals [closed]

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I am completely clueless about the approach to solve this question. I tried applying l ho'pital rule but am getting stuck.

Kindly help.

asked Nov 19 at 19:27

Akash Gautama

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closed as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil Nov 20 at 2:55

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up vote
-2
down vote

favorite

enter image description here

I am completely clueless about the approach to solve this question. I tried applying l ho'pital rule but am getting stuck.

Kindly help.

asked Nov 19 at 19:27

Akash Gautama

915

closed as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil Nov 20 at 2:55

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." – José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil

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

add a comment |

up vote
-2
down vote

favorite

enter image description here

I am completely clueless about the approach to solve this question. I tried applying l ho'pital rule but am getting stuck.

Kindly help.

asked Nov 19 at 19:27

Akash Gautama

915

enter image description here

I am completely clueless about the approach to solve this question. I tried applying l ho'pital rule but am getting stuck.

Kindly help.

limits

asked Nov 19 at 19:27

Akash Gautama

915

asked Nov 19 at 19:27

Akash Gautama

915

asked Nov 19 at 19:27

Akash Gautama

915

asked Nov 19 at 19:27

Akash Gautama

915

asked Nov 19 at 19:27

Akash Gautama

915

closed as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil Nov 20 at 2:55

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." – José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil

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

closed as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil Nov 20 at 2:55

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." – José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil

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

add a comment |

1 Answer
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HINT:

Write $frac1{(1+x^2)^n} =e^{-nlog(1+x^2)}$.

Then note that all of the contribution to the value of the limit of interest comes from the integration around $0$.

So, heuristically we can approximate the integral of interest with its lower bound

$$sqrt n int_0^1 e^{-nlog(1+x)^2},dx approx sqrt n int_0^1 e^{-n x^2},dx$$

Can you finish now?

Alternatively, to be rigorous, just let $x=y/sqrt n$ in the original integral and apply the Dominated Convergence Theorem.

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

HINT:

Write $frac1{(1+x^2)^n} =e^{-nlog(1+x^2)}$.

Then note that all of the contribution to the value of the limit of interest comes from the integration around $0$.

So, heuristically we can approximate the integral of interest with its lower bound

$$sqrt n int_0^1 e^{-nlog(1+x)^2},dx approx sqrt n int_0^1 e^{-n x^2},dx$$

Can you finish now?

Alternatively, to be rigorous, just let $x=y/sqrt n$ in the original integral and apply the Dominated Convergence Theorem.

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

add a comment |

up vote
1
down vote

HINT:

Write $frac1{(1+x^2)^n} =e^{-nlog(1+x^2)}$.

Then note that all of the contribution to the value of the limit of interest comes from the integration around $0$.

So, heuristically we can approximate the integral of interest with its lower bound

$$sqrt n int_0^1 e^{-nlog(1+x)^2},dx approx sqrt n int_0^1 e^{-n x^2},dx$$

Can you finish now?

Alternatively, to be rigorous, just let $x=y/sqrt n$ in the original integral and apply the Dominated Convergence Theorem.

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

add a comment |

up vote
1
down vote

HINT:

Write $frac1{(1+x^2)^n} =e^{-nlog(1+x^2)}$.

Then note that all of the contribution to the value of the limit of interest comes from the integration around $0$.

So, heuristically we can approximate the integral of interest with its lower bound

$$sqrt n int_0^1 e^{-nlog(1+x)^2},dx approx sqrt n int_0^1 e^{-n x^2},dx$$

Can you finish now?

Alternatively, to be rigorous, just let $x=y/sqrt n$ in the original integral and apply the Dominated Convergence Theorem.

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

HINT:

Write $frac1{(1+x^2)^n} =e^{-nlog(1+x^2)}$.

Then note that all of the contribution to the value of the limit of interest comes from the integration around $0$.

So, heuristically we can approximate the integral of interest with its lower bound

$$sqrt n int_0^1 e^{-nlog(1+x)^2},dx approx sqrt n int_0^1 e^{-n x^2},dx$$

Can you finish now?

Alternatively, to be rigorous, just let $x=y/sqrt n$ in the original integral and apply the Dominated Convergence Theorem.

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

edited Nov 19 at 21:00

answered Nov 19 at 20:45

Mark Viola

129k1273170

answered Nov 19 at 20:45

Mark Viola

129k1273170

answered Nov 19 at 20:45

Mark Viola

129k1273170

add a comment |

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