Sequence of functions convergence Define the sequence $g_n(x) = {x^n}({1-x)}.$ Does it converge uniformly on $[0,1]$?
I've been thinking about looking at its derivative, but am not too sure how to proceed. Thanks in advance for your help.

$g_n(x)$ is always nonnegative on $[0,1]$.  Its logarithmic derivative is $\frac nx - \frac{1}{1-x}$, which is positive  if $0<x<n/(n+1)$, negative if $n/(n+1)<x<1$, and zero if $x=n/(n+1)$.  So, $g_n(x)$ has its maximum at $x=n/(n+1)$, where its value is
$$
g_n(\frac{n}{n+1})=(1-\frac{1}{n+1})^n\frac{1}{n+1}\le \frac{1}{n+1}.
$$
Therefore, $g_n(x)$ converges uniformly to $0$ on $[0,1]$.
