Integration inside limit

Integrating by parts gives $I(n) \; = \; \int_0^1 (n+1)x^n \sin(\pi x^2) \cos(\pi x)\,dx \; =\; \int_0^1\big(\pi \sin(\pi x^2)\sin(\pi x) - 2\pi x\cos(\pi x^2)\cos(\pi x)\big)x^{n+1}\,dx$ Given that $\pi\sin(\pi x^2)\sin(\pi x) - 2\pi x\cos(\pi x^2)\cos(\pi x)$ is a continuous, so bounded, function on $[0,1]$ , the Dominated Convergence Theorem tells us that $I(n) \to \boxed{0}$ as $n \to \infty$ .

We have when $f \in C[0,1] , \lim_ {n \to \infty} \int_{0}^1 (n+1) x^n f(x)\ dx = f(1)$ . Proof -1

By Weierstrass approximation theorem we have there exists a $m^{th}$ order polynomial $P = a_m x^m + a_{m-1} x^{m-1} +...+ a_1 x+a_0$ such that $f \sim P$ . Now you integrate $x^n P$ and find the limit.

Proof-2 Please click this link Stack exchange answer . This is more rigorous and it uses the fact that $f$ is uniformly continuous and bounded in $[0,1]$ .