Limit of Integral of Continuous Function times monomoial

Question

Let $f$ be continuous on $[0,1]$. Find the limit:

$$ \lim_{n\to \infty} (n+1) \int_0^1 x^n f(x) \mathrm{d}x $$ I know the integral converges because $\int x^n \mathrm{d}x = \frac{x^{n+1}}{n+1} + C$. However, it seems like what the integral converges to is contingent on $f$. Does anyone have any ideas?

Intuitively speaking, it seems like the integral should tend to $0$, as $x^{n}$ decays to $0$ on the set $[0,1)$ as $n \to \infty$. How can I rigorize this? On second thought, it seems like it should tend to $f(1)$. The $n+1 \to \infty$ and this occurs at the point $x = 1$ (all the other points decay). Hence, we get a sort of dirac delta/unit impulse at $x = 1$ for large $n$.

Answer 1

If we substitute $x^n=y$ and use Lebesgue's dominated convergence theorem we get \begin{equation*} (n+1)\int_{0}^{1}x^{n}f(x)\, \mathrm{d}x = \dfrac{n+1}{n}\int_{0}^{1}y^{1/n}f(y^{1/n})\, \mathrm{d}y \to \int_{0}^{1}f(1)\, \mathrm{d}y =f(1), n\to\infty . \end{equation*}

Answer 2

For any $\epsilon>0$, there exists a number $\delta>0$ such that $|f(x)-f(1)|<\epsilon$, whenever $1-\delta<x\le 1$. Then, we can write

$$\begin{align} \left|(n+1)\int_0^1 x^n(f(x)-f(1))\,dx\right|&\le\left|(n+1)\int_0^{1-\delta} x^n(f(x)-f(1))\,dx\right|\\\\ &+\left|(n+1)\int_{1-\delta}^1 x^n(f(x)-f(1))\,dx\right|\\\\ &\le (n+1)\int_0^{1-\delta} x^n|f(x)-f(1)|\,dx\\\\ &+(n+1)\int_{1-\delta}^1 x^n|f(x)-f(1)|\,dx\\\\ &\le 2\sup_{0\le x\le 1}(f(x))\, (1-\delta)^{n+1}+\epsilon (1-(1-\delta)^{n+1})\tag1 \end{align}$$

Letting $n\to \infty$ in $(1)$ yields for any given $\epsilon>0$

$$\lim_{n\to\infty}\left|(n+1)\int_0^1 x^n(f(x)-f(1))\,dx\right|\le \epsilon$$

And we are done! The limit of interest is $f(1)$.