Limit of an integral $\int_0^{\pi/2} \frac{x\sin^nx}{\sqrt{1+\sin^2 x}}\ dx$

Question

What's the limit of $$\lim_{n\longrightarrow \infty} \int_0^{\pi/2} \frac{x\sin^nx}{\sqrt{1+\sin^2 x}}\ dx\ ?$$ Lebesgue's theorem is useless here I think, since $\sin x$ has no limit for $x\in[0,\pi/2]$ So, what do you propose?

Answer 1

Note that for any $x\in [0,\pi/2)$ : $$\left| \frac{x \sin^n x}{\sqrt{1+\sin^2 x}} \right|\leq x\leq \frac{\pi}{2} $$ Therefore, we can apply the dominated convergence theomrem to get : $$\lim_{n\to \infty} \int_0^{\pi/2} \frac{x \sin^n x}{\sqrt{1+\sin^2 x}} \ \mathrm{d}x =\int_0^{\pi/2} \lim_{n\to \infty} \frac{x \sin^n x}{\sqrt{1+\sin^2 x}} \ \mathrm{d}x =\int_0^{\pi/2} 0\ \mathrm{d}x=0$$ the limit is zero because for $x\in [0,\pi/2)$ we have $\sin^n x\to 0$ pointwisely, the case $x=\pi/2$ is negligeable since the measure of a singleton is $0$.

Answer 2

$\lim_{n\to\infty } \sin^n x =\begin{cases} 0 \mbox{ for } 0\leq x <\frac{\pi}{2} \\ 1 \mbox{ for } x=\frac{\pi}{2} \end{cases}$