Prove that $f(x)x^n$ converges to $0$ uniformly on $[0,1]$. Suppose $f: [0,1] \to \mathbf{R}$, $f$ is continuous, and $f(1) = 0$. Prove that $f(x)x^n$ converges to $0$ uniformly, as $n \to \infty$, on $[0,1]$.
We need to show $\forall \epsilon > 0, \exists N_\epsilon$ so that $n > N_\epsilon \Rightarrow |f(x)x^n| < \epsilon,$ for all $x \in [0,1]$.
I've tried graphing a few examples and see that when $|f(x)| < 1$, the function $f(x)x^n$ appears to converge very quickly to $0$, but when it's greater than $1$ it can take awhile. How do I go about proving uniform convergence?
 A: Hint: Just to show you how to approach the problem, let us try to show uniform convergence of $f(x)x^n$ to zero on the interval $[0,\delta]$ when $\delta < 1$. Since $f$ is continuous on $[0,\delta]$, it is bounded on this interval by say $M$. Then
$$ |f(x)x^n| \leq Mx^n \leq M\delta^n $$
so if we choose $n$ such that $M\delta^n < \varepsilon$, we are done.
Note that we didn't show uniform convergence on $[0,1]$ but we also haven't used the fact that $f(1) = 0$. The problem is that when $x = 1$, the term $x^n$ is $1$ so without this information, we can't even deduce that $f(x) x^n$ converges pointwise to $0$ (because at $x = 1$, it converges to $f(1)$). 
Can you see how to adapt the previous argument to show uniform convergence on the whole of $[0,1]$ using the fact that $f$ is continuous at $x = 1$ and $f(1) = 0$?
A: By continuity of $f$ we have $\vert x^nf(x) \vert < \epsilon$ for all $n$ on some interval $[1 - \delta, 1]$. Then
$$\vert x^n f(x) \vert \leq (1 - \delta)^n M$$
on $[0,1-\delta]$, where $M$ is the maximum of $f$ on $[0,1]$ (such a maximum exists by continuity).
