Evaluate $\lim_{n \to \infty} \int_0^1 \frac{dx}{x^n + 1}$ 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 $0 \leq x \leq 1$?
 A: 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$$\lim_{n\to\infty}\int_0^1\frac1{x^n+1}\,\mathrm dx=1.$$
A: 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.
