Prove $\lim_{n\to \infty} n\int_0^1x^nf(x) \,\text dx=0 $ Assume $f(x)$ is continuous on $[0,1]$ , and $f(1)=0$. Prove
$$\lim_{n\to \infty} n\int_0^1x^nf(x)\,\text dx=0 $$
I already know that $\lim_{n\to \infty}\int_0^1x^nf(x) \, \text dx=0$. Is this helpful in the question above?

Fix some $\epsilon>0$. Then there is a $\delta>0$ (smaller one) so that on the interval $[1-\delta,1]$ we have $|f|<\epsilon$.
Now we can easily estimate:
$$
\begin{aligned}
0
&\le
\left|n\int_0^1 x^n\; f(x)\; dx\right|
\\
&\le
\int_0^1 (n+1)x^n\; |f(x)|\; dx
\\
&=
\int_0^{1-\delta}(n+1)x^n\; |f(x)|\; dx
+
\int_{1-\delta}^1(n+1)x^n\; |f(x)|\; dx
\\
&\le
\int_0^{1-\delta}(n+1)x^n\; \|f\|\; dx
+
\int_{1-\delta}^1(n+1)x^n\; \epsilon\; dx
\\
&
\le 
(1-\delta)^{n+1} \|f\|
+
\epsilon
\ .
\end{aligned}
$$
We pass to the limit (superior) w.r.t. $n$ now in the obtained inequality, getting 
$$
\limsup_n\left|n\int_0^1 x^n\; f(x)\; dx\right|\le \epsilon\ .
$$
Now we let $\epsilon$ go to zero. So the limit exists, and is zero.
