Limit involving integrals [on hold]
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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.
limits
put on hold as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil 2 days ago
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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.
limits
put on hold as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil 2 days ago
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.
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up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
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
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
limits
asked 2 days ago
Akash Gautama
915
915
put on hold as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil 2 days ago
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.
put on hold as off-topic by José Carlos Santos, amWhy, RRL, TheGeekGreek, Chris Godsil 2 days ago
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.
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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.
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1 Answer
1
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1 Answer
1
active
oldest
votes
active
oldest
votes
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.
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.
add a comment |
up vote
1
down vote
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.
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 2 days ago
answered 2 days ago
Mark Viola
129k1273170
129k1273170
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