Looking for a hint on the following integration problem 
Let f(x) be continuous on [0,1]. Calculate $\lim_{n \to \infty} n\int_0^1 x^n f(x)dx$. 

What immediately jumps out at me is how close $\frac{x^nf(x)}{\frac{1}{n}}$ looks to a derivative, i.e. if I only had the function $$\frac{(x+\frac{1}{n})^nf(x+\frac{1}{n})-x^nf(x)}{\frac{1}{n}}$$ inside the integrand, then I could answer the question with $1^nf(1)-0^nf(0)=f(1)$. So I am at a dead end down this path. 
I've also tried using the Lebesgue DCT by wondering whether the series of functions $nx^nf(x)\to 0$ pointwise on $[0,1)$ since $f(x)$ is continuous and thus bounded on $[0,1]$. 
Another attempt involved trying to use the second MVT, but that doesn't succeed, either, since I do not know beforehand that the value $k \in [0,1]$ such that $$n\int_0^1 x^n f(x)dx=f(k)\int_0^k nx^ndx$$ is actually $k=1$. Any hint would be appreciated! 
 A: Estimate $\int_0^1 nx^n \left(f(x)-f(1)\right) dx$ when $n \to \infty$, using the preceding hints.
A: Here are some hints: $\int_0^1 nx^n\,dx\to1$ and $\int_0^a nx^n\,dx\to0$ as $n\to\infty$, for any $0\le a<1$.
A: Alternative solution (if you have learned about functionals):
Set $\{x^k | k\geq 0\}$ is a fundamental set in $C[0, 1]$. 
If we define functional $\Lambda_n(f)=n\int_0^1 x^n f(x) dx$ for every $n\in\mathbb N$, it is easy to calculate $\lim_{n\to\infty}\Lambda_n(x^k)=\lim_{n\to\infty} n\int_0^1 x^n x^k dx=\lim_{n\to\infty} \frac{n}{n+k+1}=1$. 
(Notice: $\Lambda_n$ is corectly defined, linear and bounded: $\|\Lambda_n\|\leq 1$ for all $n\in\mathbb N$).  
Therefore,
$\Lambda_n(a_k x^k+\dots+a_1 x+a_0)=a_k+\dots+a_1+a_0$. 
If we define functional $\Lambda(f)=f(1)$ (which is linear and bounded!) , it is easy to conclude that $\lim_{n\to\infty}\Lambda_n(f)=\Lambda(f)$ on fundamental set. 
Therefore, by $Banach-Steinhaus$ theorem we can conclude that $\lim_{n\to\infty}\Lambda_n(f)=\Lambda(f)$ on $C[0,1]$, or $\lim_{n\to\infty} n\int_0^1 x^n f(x) dx=f(1)$, for every $f\in C[0,1]$.
A: Here's a rough outline of one proof (the same as suggested above, essentially):


*

*Let $\epsilon>0$.

*Choose $1>\delta>0$ so that $f(x)<f(1)+\epsilon$ for all $x\in[\delta,1]$.

*Write
$$
\int_0^1 n x^n f(x)\,dx=
\underbrace{\int_0^\delta n x^n f(x)\,dx}_A+
\underbrace{\int_\delta^1 n x^n f(x)\,dx}_B
$$

*Using the fact that $|f|$ is bounded on $[0,\delta]$, show that $\lim\limits_{n\rightarrow\infty} |A|=0$.

*Show that $\limsup\limits_{n\rightarrow\infty}B\le f(1)+\epsilon$.

*As $\epsilon$ was arbitrary, use 3., 4., and 5. to infer that $$\limsup\limits_{n\rightarrow\infty}\int_0^1 nx^nf(x)\,dx\le f(1).$$

*In a similar manner, show that 
$$\liminf\limits_{n\rightarrow\infty}\int_0^1 nx^nf(x)\,dx\ge f(1).$$

