Evaluate $\lim_{n\to\infty} \sqrt{n}\int_{0}^{\pi/2} \sin^{n} x dx$ Question:
Compute $I_n=\lim\limits_{n\to\infty} \sqrt{n}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx$.
Attempt:
We know the famous result that $\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=0$,
and by dividing it into two parts $\int_{0}^{\frac{\pi}{2}-n^{-\alpha}}+\int_{\frac{\pi}{2}-n^{-\alpha}}^{\frac{\pi}{2}}(0<\alpha<\frac{1}{2})$ ,
we can show that $\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=O(\frac{1}{n^{\alpha}})$.
There is a hint which says by substitution we can rewrite the limits with $\Gamma$, and eventually get $I_n\sim c\frac{1}{\sqrt{n}},c>0$.
Is there any method without using the $\Gamma$ function? (Answers using Gamma function are also appreciated.)

Proceeding along your approach, without using the gamma function
Let
\begin{align*}
 J_n &:= \sqrt{n}\int_{\pi/2 - n^{-1/3}}^{\pi/2} \sin^n x \,\mathrm{d} x, \\
 K_n &:= \sqrt{n}\int_0^{\pi/2 - n^{-1/3}} \sin^n x \,\mathrm{d} x.
\end{align*}
We have
$$J_n
= \sqrt{n} \int_0^{n^{-1/3}} \cos^n y \,\mathrm{d} y
= \int_0^{n^{1/6}} \cos^n \frac{z}{\sqrt n}\, \mathrm{d} z
= \int_0^{n^{1/6}}
\mathrm{e}^{n\ln \cos \frac{z}{\sqrt n}}\mathrm{d} z.$$
Fact 1: For $0 \le u \le 1$, it holds that
$$-\frac{u^2}{2} - u^4 \le \ln \cos u \le - \frac{u^2}{2}.$$
(The proof is not difficult. Omitted.)
Using Fact 1, we have, for all $0 \le z \le n^{1/6}$,
$$-\frac{z^2}{2} - \frac{(n^{1/6})^4}{n}\le -\frac{z^2}{2} - \frac{z^4}{n} \le n\ln \cos \frac{z}{\sqrt n} \le -\frac{z^2}{2}.$$
Thus, we have
$$\mathrm{e}^{-n^{-1/3}}\int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z \le J_n \le \int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z.$$
Thus,
$$\lim_{n\to \infty} J_n = \int_0^\infty \mathrm{e}^{-z^2/2}\mathrm{d} z = \sqrt{\pi/2}.$$
$\phantom{2}$
On the other hand, we have
\begin{align*}
 K_n 
 &\le  \sqrt{n}\, \cdot \frac{\pi}{2} \sin^n \left(\pi/2 - n^{-1/3}\right)\\
 &= \frac{\pi}{2} \sqrt{n}\, \cos^n \frac{1}{\sqrt[3]{n}}\\
 &\le \frac{\pi}{2} \sqrt{n}\,
 \left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n
\end{align*}
where we have used $\cos y \le 1 - \frac{y^2}{4}$
for all $y\in [0, \pi/2]$ (easy). Also,
$$\lim_{n\to \infty} \frac{\pi}{2} \sqrt{n}\,
\left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n = 0.$$
Thus,
$$\lim_{n\to \infty} K_n  = 0.$$
$\phantom{2}$
Finally, we have
$$\lim_{n\to \infty} I_n = \lim_{n\to \infty} J_n + \lim_{n\to \infty} K_n = \sqrt{\pi/2}.$$
