$$\lim_{n \to \infty} \int_0^{\pi/2}\sin^{n}(x)dx=0$$
First I tried by recurrence relation
$$\int_0^{\pi/2}\sin^{n}(x)dx=I_n=\frac{n-1}{n} I_{n-2}$$
Than $I_n$ is cauchy sequence because $\frac{n-1}{n}<1$ So it's bounded and converge.
But how to prove limit is 0?
Futhermore, It is possible that prove without using recurrence relation?
This is an immediate consequence of DCT (Dominated Convergence Theorem).
Ref . https://en.wikipedia.org/wiki/Dominated_convergence_theorem
Proof without using DCT. Let $\epsilon >0$. Note that $\int_{\pi /2-\epsilon}^{\pi/2} \sin^{n} x dx \leq \int_{\pi /2-\epsilon}^{\pi/2} 1 dx=\epsilon$. Now consider $\int_0^{\pi /2-\epsilon} \sin x^{n} dx$. Observe that the supremum of $\sin x$ on $[0,\pi /2-\epsilon]$ is a number $r$ less than $1$. Hence $\int_0^{\pi /2-\epsilon} \sin x^{n} dx \leq r^{n} (\pi /2) \to 0$. Now just add the two integrals.