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.

We can prove that if $f$ is continuous function ($f$ being differentiable function is not needed)
$$
\lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx=f(1)
$$
Since $f$ is continuous, $f$ is bounded on $[0,1]$. Also
$$
\forall \epsilon>0, \exists \delta>0, \forall x\in(1-\delta,1],\text{there is } |f(x)-f(1)|<\epsilon 
$$
\begin{align}
\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|&=
\left|\int_{0}^{1-\delta} (n+1)x^nf(x)dx+\int_{1-\delta}^{1} (n+1)x^nf(x)dx-f(1)\right|
\\
&=\left|f(t_1) \int_{0}^{1-\delta} (n+1)x^ndx+f(t_2)\int_{1-\delta}^{1} (n+1)x^ndx-f(1)\right| \hspace{5 mm}  
\\ &\hspace{10 mm}(t_1\in(0,1-\delta),t_2\in(1-\delta,1) \text{ and by IMVT})
\\
&=\left|f(t_1)(1-\delta)^{n+1}+f(t_2)(1-(1-\delta)^{n+1})-f(1)\right|
\\
&\leqslant 2M(1-\delta)^{n+1}+|f(t_2)-f(1)|
\\&<2M(1-\delta)^{n+1}+\epsilon
\end{align}
So
$$
\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|\leqslant \varlimsup\limits_{n\to\infty}2M(1-\delta)^{n+1}+\epsilon=\epsilon
$$
Since $\epsilon$ is arbitrary small
$$
\varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 \hspace{5 mm} \text{and}\hspace{5 mm} \varliminf\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 
$$
So 
$$
\lim\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0\hspace{5 mm} \text{or} \hspace{5 mm}\lim\limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)
$$
