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Is the following proof adequate to evaluate; $$\lim_{n \to \infty} \int_0^1 \frac{dx}{x^n + 1}$$

Proof

As the upper and lower bounds of integration are respectively $0$ and $1$, we have that $0<x<1, x\in \mathbb{R}$, and hence, as $|x|<1 \Rightarrow \lim_{n \to \infty} |x|^n = 0 \Rightarrow \lim_{n \to \infty} x^n=0$. Consequently; $$\lim_{n \to \infty} \int_0^1 \frac{dx}{x^n + 1} = \int_0^1 \lim_{n \to \infty} \frac{dx}{x^n + 1} = \int_0^1 \frac{dx}{0+1} = 1$$

Issue

One of my issues with the proof is that it seems to exclude the upper bound when $x=1$, do the bounds of integration restrict the values of $x$ to $0<x<1$ or to $0 \leq x \leq 1$?

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    $\begingroup$ Interchanging integrals and limits isn't valid in general (Googling "gliding hump" will probably lead you to examples). Though you could apply the Dominated Convergence Theorem to justify the interchange in this case. $\endgroup$ – Daniel Schepler Jan 12 '18 at 21:44
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You can't just switch like that the limit and the integral! However, it is true that the limite is $1$. Take $\varepsilon\in(0,1)$. If $n$ is big enough, then $$(\forall x\in[0,1-\varepsilon]):\frac{x^n}{x^n+1}<\varepsilon$$and therefore$$\int_0^1\frac{x^n}{x^n+1}\,\mathrm dx=\int_0^{1-\varepsilon}\frac{x^n}{x^n+1}\,\mathrm dx+\int_{1-\varepsilon}^1\frac{x^n}{x^n+1}\,\mathrm dx<\varepsilon(1-\varepsilon)+\frac{\varepsilon}2.$$Since this number is as small as you wish,$$\lim_{n\to\infty}\int_0^1\frac{x^n}{x^n+1}\,\mathrm dx=0.$$Since$$(\forall x\in[0,1]):\frac{x^n}{x^n+1}+\frac1{x^n+1}=1,$$this is the same thing as saying that$$\lim_{n\to\infty}\int_0^1\frac1{x^n+1}\,\mathrm dx=1.$$

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  • $\begingroup$ Does it equal $0$ or $1$? $\endgroup$ – frog1944 Jan 12 '18 at 21:50
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    $\begingroup$ @frog1944 I made a mistake. I hope that all is well now. $\endgroup$ – José Carlos Santos Jan 12 '18 at 21:57
  • $\begingroup$ $\frac{x^n}{x^n+1}\leq x^n$. $\endgroup$ – Riemann Feb 1 at 6:46
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There is a big problem with your proof. The problem is not that the point $x=1$ is left out, the problem is that in general $$\lim_{n\to\infty}\int_0^1f_n\ne\int_0^1\lim_{n\to\infty}f_n.$$That works if the sequence $(f_n)$ converges uniformly, but here the convergence is not uniform.

Hint: Let $\epsilon>0$. Since you have $0\le f_n\le 1$ you have $$\left|\int_{1-\epsilon}^1f_n\right|<\epsilon$$for every $n$. And you do have $f_n\to1$ uniformly on $[0,1-\epsilon]$, so $$\left|\int_0^{1-\epsilon}f_n-(1-\epsilon)\right|<\epsilon$$if $n$ is large enough; hence the triangle inequality shows that $$\left|\int_0^1f_n-1\right|<3\epsilon$$if $n$ is large enough.

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