How does one show that $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^{n}}\, dx = 0$?

Question

How does one show that $$\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^{n}}\, dx = 0?$$ My idea is to evaluate the inner integral, but I can't seem to be able to do that.

Answer 1

For all $x\in [0,1]$ you have $$ \frac{x^n}{1+x^n} \leq x^n $$ and hence

$$ 0\leq\int_{0}^{1}\frac{x^{n}}{1 + x^{n}}\, dx \leq \int_{0}^{1}x^{n}\, dx = \frac{1}{n+1}\rightarrow 0 \quad (n\rightarrow \infty).$$

Answer 2

If you really want to evaluate an integral, note that $$\frac{x^n}{1+x^n}\le \frac{x^{n-1}}{1+x^n}$$ on $[0,1]$ and let $u=1+x^n$.