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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.

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This follows immediately from the dominated convergence theorem, if you are familiar with that. – Potato Sep 13 '12 at 3:15
up vote 12 down vote accepted

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

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

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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 $u=1+x^n$.

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