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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 immedia$t$ely from the dominated convergence theorem, if you are familiar with that.2012-09-13

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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 $u=1+x^n$.