Limit of an integral on [0,1] I need to prove a limit on a closed interval $[0,1]$. 
$f$ is integrable.
$$\lim_{n\to \infty}\int_0^1 x^n f(x) \, dx=0$$
Where should I start from?
 A: If by "integrable" you mean Riemann integrable, then an integrable function must be bounded, and so
$$\left|\int\limits_{0}^{1}{x^nf(x)\text{ d}x}\right|\le\left(\max\limits_{x\in[0,1]}{|f(x)|}\right)\int\limits_{0}^{1}{x^n\text{ d}x}. $$
Note that $\int\limits_{0}^{1}{x^n\text{ d}x} = \frac{1}{n+1}$.
If by "integrable" you mean Lebesgue integrable, then the functions $f_n(x) = x^nf(x)$ converges pointwise to zero almost everywhere, and $|f_n(x)|\le|f(x)|$ everywhere, so we can apply the Dominated Convergence Theorem.
A: I assume you mean $f\in L^1(0,1)$. Note $C[0,1]$ is dense in $L^1(0,1)$ under $L^1$ norm, so it suffices to prove it with $f\in C[0,1]$ and then pass to limit (Please see the comments for details).
Given $f\in C[0,1]$, by uniformly continuous, for any $\epsilon>0$, you can find a step function $g$, $0\le f(x)-g(x)\le \epsilon$ for all $x\in[0,1]$, so it suffices to prove it for step functions and then pass to limit.
Then by linearity of integral, it suffices to prove it for $f=1_{[a,b]}$, which is easy by simple calculation.
A: Since $f$ is integrable, it is finite almost everywhere. Then $x^n\,f(x)$ converges to $0$ almost everywhere in$[0,1]$. Moreover $|x^n\,f(x)|\le|f(x)|$ for all $x\in[0,1]$. The result follows now from the Dominated Convergence Theorem.
