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.
 A: You may just integrate by parts:
$$
(n+1)\int_0^1x^{n}f(x)dx=\left.x^{n+1}f(x)\right|_0^1-\int_0^1x^{n+1}f'(x)dx=f(1)-\int_0^1x^{n+1}f'(x)dx
$$ and use the fact that $f'$ is continuous over $[0,1]$ giving
$$
|f'(x)|\leq M, \quad x \in [0,1],
$$ and
$$
\left|\int_0^1x^{n+1}f'(x)dx\right|\leq\int_0^1x^{n+1}|f'(x)|dx\leq\frac{M}{n+2}
$$ to conclude.
A: 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 
$$
\lim\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0\hspace{5 mm} \text{or} \hspace{5 mm}\lim\limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1)
$$
A: 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 $f(1)$.
