integral of $x^{1/n}\sin x$ how to prove $\lim_{n \to \infty} \int_0^{\pi/2} x^{1/n}\sin x dx = 1$? I've tried it by using squeeze theorm. It worked on right side by putting $\pi/2$ into $x$, but i have no idea for left side
 A: Here's an elementary approach (without using any measure theory):
Since $\int_0^\frac{\pi}{2} \sin(x)dx=1$, we have that
$$
\left| \int_0^\frac{\pi}{2} x^\frac{1}{n}\sin(x)dx-1\right|=\left|\int_0^\frac{\pi}{2}\sin(x)\left(x^\frac{1}{n}-1\right)dx\right|\le \int_0^\frac{\pi}{2}\left|x^\frac{1}{n}-1\right|dx 
$$
were the last inequality follows from $\left|\int f\right|\le \int |f|$ and $|\sin(x)|\le 1$. 
Since $x^\frac{1}{n}\ge 1 \iff x\ge 1$, we can write 
$$
\int_0^\frac{\pi}{2}\left|x^\frac{1}{n}-1\right|dx =\int_0^1 1-x^\frac{1}{n}dx+\int_1^\frac{\pi}{2}x^\frac{1}{n}-1dx
$$
$$
=1-\left[\frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} \right]_0^1+\left[\frac{x^{\frac{1}{n}+1}}{\frac{1}{n}+1} \right]_1^\frac{\pi}{2}-\frac{\pi}{2}+1
$$
$$
=1-\frac{1}{\frac{1}{n}+1}+\frac{\left(\frac{\pi}{2}\right)^{\frac{1}{n}+1}}{\frac{1}{n}+1}-\frac{1}{\frac{1}{n}+1}-\frac{\pi}{2}+1
$$
Letting $n\to\infty$, we see that the last expression tends to $0$, as desired. 
A: First let $\lim_{n \to \infty}$. Now when you have this, put it in the part you integrate, so you will get $\lim_{n \to \infty}$$ x^{1/n}$ (notice, you have n only in this part, so do limes for it). $\lim_{n \to \infty}$$ x^{1/n} =1$. Now you have  $\int_0^{\pi/2} 1*\sin x dx = 1$. If you have have a hard time to compute the last ontegral let me know. Hope this will help.
