Suppose that $f\in\mathscr{C}_{\mathbb{R}}\left([0,1]\right)$, show that \begin{equation} \lim\limits_{n \to \infty}\frac{\int_0^1 x^{n}f(x)dx}{\int_0^1 x^{n}dx}=f(1) \end{equation}
My idea is to use the Stone-Weierstrass Theorem but I'm a little lost, I will appreciate any help. Thanks. :D
We are looking for $$ \lim_{n\to +\infty}\int_{0}^{1}(n+1)x^n f(x)\,dx $$ and by the proof of the Weierstrass approximation theorem through Bernstein polynomials we know that for any $\varepsilon > 0$ there is some $N$ such that $$ p(x) = \sum_{k=0}^{N}f\left(\frac{k}{N}\right)\binom{N}{k}x^k(1-x)^{N-k} $$ differs at most by $\varepsilon$ from $f(x)$. Let us fix this $\varepsilon$ and the associated $N$. We may notice that for any $n$
$$ \int_{0}^{1} (n+1) x^n p(x)\,dx = \sum_{k=0}^{N}f\left(\frac{k}{N}\right)\binom{N}{k}(n+1)\int_{0}^{1}x^{n+k}(1-x)^{N-k}\,dx $$
equals
$$ \frac{n+1}{n+N+1}f(1)+\frac{n+1}{n+N+1}\sum_{k=0}^{N-1}f\left(\frac{k}{N}\right)\frac{\binom{N}{k}}{\binom{N+n}{k+n}}.$$
By sending $n$ to $+\infty$ we have that the initial limit exists and it is at most $\varepsilon$-apart from $f(1)$.
Since $\varepsilon$ is arbitrary such limit equals $f(1)$.
