Does $f_{n}(x)=n\cos^n x \sin x$ uniformly converge for $x \in [0,\frac{\pi}{2}]$? 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.
 A: We use the following claim:

Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$  which converges uniformly to $f$ on $[a,b]$. Then 
  $$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$

Indeed, we have 
$$\left|\int_a^bf_n(t)dt-\int_a^bf(t)dt\right|\leq (b-a)\sup_{a\leq x\leq b}|f_n(x)-f(x)|,$$
which converges to $0$ thanks to the uniform convergence on $[a,b]$.
In our case, we have 
$$\int_0^{\frac{\pi}2}f_n(x)dx=n\left[-\frac{(\cos x)^{n+1}}{n+1}\right]_0^{\frac \pi 2}=\frac n{n+1}\to 1,$$
whereas $f_n$ converges pointwise to $0$. This shows that the convergence cannot be uniform.
A: Let $x_n=\sin^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$. Then $\sin(x_n)=\frac{1}{\sqrt{n+1}}$ and $\cos(x_n)=\frac{1}{\sqrt{1+1/n}}$.
Thus,
$$
\begin{align}
\lim_{n\to\infty}\frac{f_n(x_n)}{\sqrt{n}}
&=\lim_{n\to\infty}\frac{n\;\cos^n(x_n)\sin(x_n)}{\sqrt{n}}\\
&=\lim_{n\to\infty}\frac{1}{(1+1/n)^{n/2}}\sqrt{\frac{n}{n+1}}\\
&=e^{-1/2}\tag{1}
\end{align}
$$
Therefore,
$$
f_n(x_n)\sim e^{-1/2}\sqrt{n}\tag{2}
$$
The asymptotic growth in $(2)$ says that, although $\lim\limits_{n\to\infty}f_n(x)=0$ pointwise, $f_n(x)$ does not converge uniformly to $0$ on $[0,\frac{\pi}{2}]$ since there is always an $x$ so that $f_n(x)>1$.
A: 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. 
