TITLE: find $\lim\limits_{n \rightarrow \infty }(n+1)\int\limits_{0}^{1} x^n f(x)$
QUESTION [6 upvotes]: Let $f(x)$ , $x \in [0,1]$,be any positive real valued continuous function. Then find $$\lim\limits_{n \rightarrow \infty }(n+1)\int\limits_{0}^{1} x^n f(x)$$
I tried by integrating by parts to get the result $f(1)-(n+1)\int\limits_{0}^{1} x^{n+1}f(x) dx$ But i think this does not help us anyway to find the limit.

REPLY [6 votes]: Notice that for any $ 0 < \delta <1$ :
$$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1-\delta} x^n f(x) = 0$$
since it is bounded by $$\ \ (n+1)\sup |f| \cdot (1-\delta)^n.$$
Therefore, for any $\delta >0 $
$$\lim_{n \rightarrow \infty }(n+1)\int_{0}^{1} x^n f(x) = \lim_{n \rightarrow \infty }(n+1)\int_{1-\delta}^{1} x^n f(x) \ .$$
Now for any $n$,
$$ \int_{1-\delta}^{1} (n+1) x^n f(x) dx 
\leq \int_{1-\delta}^{1} (n+1) x^n \sup_{\{1 - \delta \leq y \leq 1 \}} f(y) \ dx 
\leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx $$
Similarly,
$$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \int_{1-\delta}^{1} (n+1) x^n dx \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \ .$$
Now, letting $n \to \infty$, we get
$$ [ \inf_{\{1 - \delta \leq y \leq 1 \}} f(y)] \leq \int_{1-\delta}^{1} (n+1) x^n f(x) dx \leq [ \sup_{\{1 - \delta \leq y \leq 1 \}} f(y)] \ .$$
This is true for all $\delta$. So, let $\delta \to 0$. Both sides converge to $f(1)$ by contiuity of $f$ at $x=1$. End.
NOTE: This proof shows that continuity at $x=1$ and boundedness of $f$ are sufficient to claim the same argument!