TITLE: Does $f_{n}(x)=n\cos^n x \sin x$ uniformly converge for $x \in [0,\frac{\pi}{2}]$?
QUESTION [11 upvotes]: I want to check whether the following function is uniformly converges:
$f_n(x)=n\cos^nx\sin x$ for $x \in \left[0,\frac{\pi}{2} \right]$.
I proved that the $\lim \limits_{n \to \infty}f_{n}(x)=0$ for every $x$. I'd love your help with the uniformly continues convergence.  I always get confused with it.  I already showed that $|f_n(x) - 0|< \epsilon$. What else should I show or how should I refute the claim? 
Thanks a lot.

REPLY [3 votes]: You might like to simplify first (otherwise the argument is that of robjohn):
Put $t=\cos(x)$ ($0\le t\le1$), then we look at 
$$g(t)= n t^n\sqrt{1-t^2}$$
or, even better, we might consider $h(t) = g(t)^2= n^2 t^{2n}(1-t^2)$. Then 
$$h'(t)= 2n^3t^{2n-1}(1-t^2)-2n^2t^{2n+1}=2n^2t^{2n-1}(n-(n+1)t^2).$$
Note from this that $t_n=\sqrt{\frac{n}{n+1}}$ is the maximum of $h$, and that 
$$h(t_n)=n^2\left(\frac{n}{n+1}\right)^n \left(1-\frac{n}{n+1}\right)=  n\cdot\frac{1}{\left(\frac{n}{n+1}\right)^n}\cdot \frac{n}{n+1}\sim n\cdot\frac{1}{e}\cdot 1 \qquad \text{as $n\to\infty$}$$
Which should have been $0$ if the limit was uniform.