To evaluate limit $\lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx$ 
Let $f:\Bbb{R} \rightarrow \Bbb{R}$ be a differentiable function whose derivative is continuous, then
$$\lim_{n \to \infty} \left((n+1)\int_0^1x^{n}f(x)dx\right).$$

I think I have to use L'Hop rule, but I don't see how.

By condition and integration by parts formula, it follows that
\begin{align}
& (n + 1)\int_0^1 x^n f(x) dx \\
= & \int_0^1 f(x) d(x^{n + 1}) \\
= & \left.f(x)x^{n + 1} \right|_0^1 - \int_0^1 x^{n + 1}f'(x)dx \\
= & f(1) - \int_0^1 x^{n + 1}f'(x) dx.
\end{align}
Since $f'$ is continuous, $f'$ is bounded on the interval $[0, 1]$, assume it is bounded by $M \geq 0$, then,
\begin{align}
& \left|\int_0^1 x^{n + 1}f'(x) dx\right| \\
\leq & \int_0^1 x^{n + 1}|f'(x)| dx \\
\leq & M\int_0^1 x^{n + 1} dx \\
= & \frac{M}{n + 2} \to 0 
\end{align}
as $n \to \infty$.
Hence the result of the original limit is $f(1)$.
