<s>[INST] Evaluate limit integrate: $\lim_{n \rightarrow +\infty}{\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx}$ I know,
when n is even,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m-1)!!}{(2m)!!}\frac {\pi}{2} $$
and n is odd,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m)!!}{(2m+1)!!} $$
But when let limit n to infinite, how can I get the limit, or if the limit exists.
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I find a good way to solve the limit:$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!}
$$
as we know,
$2 =\frac{1+3}{2} > \sqrt{1 \cdot 3}$ , so $2n = \frac{(2n-1)+(2n+1)}{2} > \sqrt{(2n-1)(2n+1)}$
so,
$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!} = \frac{\sqrt{1 \cdot 3}\cdots \sqrt{(2n-3)(2n-1)}\sqrt{2n-1}}{\sqrt{1\cdot 3} \cdots \sqrt{(2n-3)(2n-1)} \sqrt{(2n-1)(2n+1)}} = \frac{1}{\sqrt{2n+1}} \to 0 (n \to \infty)$$
 [/INST] It is obviously zero. Think of the graph of $\sin^nx$ with $n$ going to infinity. How the area under the $y=\sin^nx $ changes accordingly
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