Finding a limit including an improper integral: $\lim_{n\to\infty}⁡ \int_0^{\sqrt n} \left({1-\frac{x^2}{n}}\right)^n\,dx$

Asked Jan 24 '18 at 19:37

Active Dec 01 '19 at 07:59

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I was given a task to calculate that limit, though I'm having trouble with it.. I thought about that fact that $(1-x^2/n)^n$ converges to $e^{-x^2}$ and about switching limit and integral when the convergence is uniform, though I can't really get it done properly. The limit is:

$$\lim_{n\to\infty}⁡ \int_0^{\sqrt n} \left({1-\frac{x^2}{n}}\right)^n\,dx$$

Any help would be truely appreciated!

edited Dec 01 '19 at 07:59

Martin Sleziak

53,687

asked Jan 24 '18 at 19:37

Tamir Shalev

Do you know the Dominated Convergence Theorem? – JimmyK4542 Jan 24 '18 at 19:50
I'm afraid not. I'm dealing with improper integrals right now and with function series (Pointwise/uniform convergences) etc.. Calculus II – Tamir Shalev Jan 24 '18 at 19:53
The sequence is monotonically increasing. Does that give you an idea how to proceed? – Daniel Fischer Jan 24 '18 at 20:05
Do you mean dini's theorem? – Tamir Shalev Jan 24 '18 at 21:14
In part. The monotonicity yields the existence of the limit, $\int_0^{\infty} e^{-x^2},dx$ gives an upper bound. Then Dini's theorem can be used to deduce that the limit is $\geqslant \int_0^a e^{-x^2},dx$ for every $a \in [0,\infty)$. – Daniel Fischer Jan 24 '18 at 22:56
Understood, thank you very much! – Tamir Shalev Jan 24 '18 at 23:21

Finding a limit including an improper integral: $\lim_{n\to\infty}⁡ \int_0^{\sqrt n} \left({1-\frac{x^2}{n}}\right)^n\,dx$

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