How to prove $\lim_{n\to \infty}\int_{-\pi/2}^{\pi/2}\cos^nxdx=0$ I need to prove that
$$\lim_{n\to \infty}\int_{-\pi/2}^{\pi/2}\cos^nxdx=0.$$
I've tried to use reduction formula, and I have got
$$\int_{-\pi/2}^{\pi/2}\cos^nxdx=k\frac{(n-1)!!}{n!!},$$
where $k=\pi$ if n is even and $k=2$ otherwise. But I can't compehend how to calculate $$\lim_{n\to \infty}k\frac{(n-1)!!}{n!!}.$$
I think there is an easier way to prove that without using this formula, but I really can't grasp it. 
 A: Note that it suffices to show $$\lim_{n\to \infty}\int_{0}^{\pi/2}\cos^nxdx=0.$$
Let $\epsilon >0$ be given. 
$$\int_{0}^{\pi/2}\cos^nxdx= \int_{0}^{\epsilon }\cos^nxdx +\int_{\epsilon }^{\pi/2}\cos^nxdx \le $$
$$\epsilon +\int_{\epsilon }^{\pi/2}\cos^nxdx  $$
Note that the function $\cos x $ is a decreasing function on the interval $[\epsilon,\pi/2]$ thus for every $x\in [\epsilon, \pi/2]$ we have $$ \cos ^n (x) \le\cos ^n(\epsilon)$$
Let $n$ be large enough that $\cos ^n (\epsilon)< \epsilon$
Then we have $$ \int_{\epsilon }^{\pi/2}\cos^nxdx \le \epsilon(\pi /2 - \epsilon)$$ 
Thus $$\int_{0}^{\pi/2}\cos^nxdx= \int_{0}^{\epsilon }\cos^nxdx +\int_{\epsilon }^{\pi/2}\cos^nxdx \le \epsilon + \epsilon ( \pi/2-\epsilon)$$
Since $\epsilon>0$ is arbitrary small, we have $$\lim_{n\to \infty}\int_{0}^{\pi/2}\cos^nxdx=0.$$
A: If you want to use your result, the explicit formula of the integral, you can apply Stirling's formula to deduce that the limit is zero. We may separate into two cases: when $n$ is even or odd. We may assume that $n=2m$ is even (and the odd case is similar). In that case, the integral is 
$$
I_{n} = \pi \frac{(n-1)!!}{n!!} = \pi \frac{(2m-1)(2m-3)\cdots 1}{(2m)(2m-2)\cdots 2} = \pi \frac{(2m)!}{2^{2m}(m!)^{2}}
$$
Now Striling's formula is
$$
m! \sim \sqrt{2\pi m}\frac{m^{m}}{e^{m}}, 
$$
so this gives
$$
I_{2m} \sim \pi \frac{\sqrt{4\pi m}\frac{(2m)^{2m}}{e^{2m}}}{2^{2m} 2\pi m \frac{m^{2m}}{e^{2m}}} = \sqrt{\frac{\pi}{m}} 
$$
as $m\to \infty$. 
A: Hint: Without that $k$ factor in front, the limit is zero for that expression involving double factorials. So now how do you deal with that factor of $k$, which depends (mildly) on $n$?
