For $n \ge 2$, define the sequence $\{x_n\}$ by $$x_n=\frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}\tan^{\frac{1}{n}}t \;dt$$ Then the sequence $\{x_n\}$ converges to ________.
My attempt: $$\lim_{n \to \infty}x_n=\frac{1}{2\pi}\lim_{n \to \infty}\int_{0}^{\frac{\pi}{2}}\tan^{\frac{1}{n}}t \;dt=\frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}\lim_{n \to \infty}\tan^{\frac{1}{n}}t \;dt=\frac{1}{2\pi}\int_{0}^{\frac{\pi}{2}}dt=0.25$$
But, taking the limit inside the integral is not always allowed (one should show that the convergence is uniform on the finite interval). How do I justify 2nd equality sign in the solution. Can someone please help? It seems to me that my solution is fine, but the justification is required. Thanks in advance.
Hint:
In the first quadrant,
$$x\le\tan x\le\frac1{\frac\pi2-x}$$ and $$\int_0^{\pi/2}x^{1/n}dx\le\int_0^{\pi/2}\tan^{1/n}x\,dx\le\int_0^{\pi/2}\frac1{(\frac\pi2-x)^{1/n}}dx.$$
The two bracketing integrals are easy.