Limit of $\lim_{n \to \infty} \int_{0}^{\frac{\pi}{6}} \sqrt{n} \cos^n \theta d\theta$ I`m trying to prove that the volume of the intersection of the $n$-dimensional euclidean unit ball with the slab $-1/2 < x_{1} < 1/2$ is greater than $0.96$ times the volume of the $n$-dimensional euclidean unit ball for large $n$.
In the middle of the computations, I got stuck on compute
$$
\lim_{n \to \infty}\,\int_{0}^{\pi/6}
\sqrt{\, n\,}\cos^{n}\left(\theta\right)\,\mathrm{d}\theta
$$
I think that its limit is $1$, but I could not prove it. Any ideas ?.
 A: For your question on the cosine integral, do you know Laplace's method? It will give you the limit.
I think there might be easier ways to go for your result though. Essentially you want to show
$$\tag 1 \frac{\int_{1/2}^1(1-x^2)^{(n-1)/2}\,dx}{\int_{0}^1(1-x^2)^{(n-1)/2}\,dx} \to 0$$
as $n\to \infty.$ I've used Fubini and a little symmetry to get it to here. Now the numerator is no more than $(3/4)^{(n-1)/2}.$ The denominator is larger than $\int_0^1 (1-x)^{(n-1)/2}\,dx = 2/(n+1).$ It follows that the quotient in $(1)$ is bounded above by
$$\frac{(3/4)^{(n-1)/2}}{2/(n+1)},$$
which $\to 0$ rapidly.
A: Since $\cos(x)=1-\frac{x^2}2+O\!\left(x^4\right)=e^{-\frac{x^2}2}\left(1+O\!\left(x^4\right)\right)$, we have that
$$
\begin{align}
\lim_{n\to\infty}\sqrt{n}\int_0^{\pi/6}\cos^n(x)\,\mathrm{d}x
&=\lim_{n\to\infty}\sqrt{n}\int_0^{\pi/6}e^{-n\frac{x^2}2}\left(1+O\!\left(nx^4\right)\right)\,\mathrm{d}x\\
&=\lim_{n\to\infty}\int_0^{\sqrt{n}\pi/6}e^{-\frac{x^2}2}\left(1+O\!\left(\frac{x^4}n\right)\right)\,\mathrm{d}x\\
&=\int_0^\infty e^{-\frac{x^2}2}\,\mathrm{d}x\\
&=\sqrt{\frac\pi2}
\end{align}
$$
A: Rewrite the integral as
$$\sqrt{n} \int_0^{\pi/6} d\theta \, e^{n \log{(\cos{\theta})}} $$
We can apply Laplace's method here.  There is a min at the origin; all contributions outside of this stationary point will be exponentially small.  Thus, as $n \to \infty$, we may approximate the integrand with a Taylor series in the exponential:
$$\sqrt{n} \int_0^{\epsilon} d\theta \, e^{-n \theta^2/2} $$
where $\epsilon$ is the size of the neighborhood outside of which we may ignore contributions to the integral.  which, with exponentially small error, may be approximated as
$$\sqrt{n} \int_0^{\infty} d\theta \, e^{-n \theta^2/2}  = \sqrt{\frac{\pi}{2}}$$
