Let $f:[0,1]\rightarrow$ $\mathbb{R}$ be continuous and $f(1)=0$. Show that $\displaystyle\lim_{n \to \infty}\int_0^1 \! x^nf(x) \, \mathrm{d}x=0.$
This is what I've done.
Let $\epsilon>0$. Split the integral into two parts.
$\displaystyle\int_0^1 \! x^nf(x) \, \mathrm{d}x$=$\displaystyle\int_0^{1-\epsilon} \! x^nf(x) \, \mathrm{d}x + \displaystyle\int_{1-\epsilon}^1 \! x^nf(x) \, \mathrm{d}x$.
Since $x^n\rightarrow0$ uniformly on $[0,1-\epsilon]$, the first integral goes to zero when we take the limit.
Now it's the second integral that's giving me trouble. I wanted to say that since f is uniformly continuous on $[0,1]$, we have that
$\displaystyle\int_{1-\epsilon}^1 \! x^nf(x) \, \mathrm{d}x=\displaystyle\int_{1-\epsilon}^1 \! x^n(f(x)-f(1)) \, \mathrm{d}x<\displaystyle\int_{1-\epsilon}^1 \! \epsilon \, \mathrm{d}x$ for any $x\in[1-\epsilon,1]$.
Is this true? Any suggestions would be helpful. Thanks!