Evaluate
$$\lim\limits_{n\to \infty}\int_{0}^{1}\frac{\sqrt{n}(e^{-x/n}-1)}{x}dx.$$
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Mhenni Benghorbal
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Hugh C.
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2 Answers
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One doesn't need the full strength of Taylor's theorem. By Mean Value Theorem, there is some $\xi \in[0,x/n]$ such that $|\frac{e^{-x/n} - 1}{x/n}| = e^{-\xi} \le 1.$ Thus $$ \sqrt{n} \frac{e^{-x/n} - 1}{x} \le \frac{1}{\sqrt n} \to 0.$$ So the integrand uniformly converges to $0$, which means that the integral is 0.
Calvin Khor
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Hint: You can interchange the limit with the integral if the sequence converges uniformly. So, can you prove that the sequence converges uniformly?
Mhenni Benghorbal
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thanks for your prompt response! I get your point. For this sequence of function should i prove $\frac{\sqrt{n}(e^{-x/n}-1)}{x}$ uniformly converges to $\frac{e^{-x}-1}{x}$? – Hugh C. Nov 20 '13 at 20:52
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Try to find the pointwise limit of $\sqrt{n}(e^{-x/n}-1)$. – Mhenni Benghorbal Nov 20 '13 at 20:55
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Take your time to work this out. See the techniques for proving uniform convergence of a sequence. – Mhenni Benghorbal Nov 20 '13 at 20:56
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@user92654: See the technique beneath "note" in my answer which helps you to find the limit. – Mhenni Benghorbal Nov 20 '13 at 21:14
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for this question, you wanna me think in the Taylor series way? – Hugh C. Nov 20 '13 at 22:25
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It is a technique to find the limit. Then see if the sequence converges uniformly to this limit for $x\in[0,1]$. – Mhenni Benghorbal Nov 20 '13 at 22:55
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I guess the sequence should unif $\to 0 for x\in [0,1]$, but i am really suffering how to prove it... – Hugh C. Nov 21 '13 at 16:57