TITLE: limit of integral $n\int_{0}^{1} x^n f(x) \text{d}x$ as $n\rightarrow \infty$
QUESTION [30 upvotes]: I am trying to solve the following problem at the level of a senior undergrad analysis level. So, the problem is as follows: We are given a function $f$ which is continuous on the interval $\left [ 0,1 \right ]$, and the question is to find the limit: $$\lim_{n\rightarrow \infty}\int_{0}^{1}x^{n}f(x)dx\;.$$ The second part of the problem is to deduce the following limit: $$\lim_{n\rightarrow \infty}n\int_{0}^{1}x^{n}f(x)dx\;.$$
For the first part: I just did the following: For every $0\leq x< 1$: $x\leq M$, where $0< M< 1$. Then: $$\int_{0}^{1}x^{n}f(x)dx\leq M^{n}\int_{0}^{1}f(x)dx\;.$$ Then: $$\lim_{n\rightarrow \infty }\int_{0}^{1}x^{n}f(x)dx\leq \lim_{n\rightarrow \infty }M^{n}\int_{0}^{1}f(x)dx=
0.\int_{0}^{1}f(x)dx=0\;,$$ so $$\lim_{n\rightarrow \infty }\int_{0}^{1}x^{n}f(x)dx=\lim_{n\rightarrow \infty }f(1)\int_{0}^{1}1dx=f(1)\;.$$ Does that make sense? If not, please show me the correct one.
As for the second part, I have no idea what to do. Any help?

REPLY [4 votes]: A different approach:
First let’s show the equality for polynomials;
Let $p=a_0+\cdots+a_mx^m$ then $$\int_0^1nx^np(x)dx=n(\frac{a_0}{n+1}+\cdot+\frac{a_m}{n+m+1})$$
So as $n\to\infty$ the RHS tends to $a_0+\cdots+a_m=p(1).$
If $f$ is a continuous function in the unit interval for the given $\epsilon>0$ Find a polynomial p so that $\|p-f\|<\epsilon$ Then
$$\Big|\int_0^1nx^nf(x)dx-f(1)\Big|=\Big|\int_0^1nx^nf(x)dx-f(1)+p(1)-p(1)-\int_0^1nx^np(x)dx+\int_0^1nx^np(x)dx\Big|\leq \Big|\int_0^1nx^n[f(x)-p(x)]dx\Big|+\Big|\int_0^1nx^np(x)dx-p(1)\Big|+|p(1)-f(1)|\leq \int_0^1n\epsilon x^ndx+\epsilon\leq \frac{n}{n+1}\epsilon+\epsilon<2\epsilon$$