Limit of the function series $f_n(x) = \sqrt{n+1}\cdot \sin^{n}(x)\cdot\cos(x)$? 
Find the limit of the function series $$f_n(x) = \sqrt{n+1}\cdot \sin^{n}(x)\cdot\cos(x).$$

The limit looks like it should be $0$, because if $\sin(x)=1$ then $\cos(x)=0$ and the function is $0$. At all other points, $|\sin(x)|=a<1$ and $\lim_{n\rightarrow\infty}a^n = 0$. But how do I know that it beats out $\lim_{n\rightarrow\infty}\sqrt{n+1}$?  
Also I was asked if the convergence happens uniformly, so I derived the function and got two critical points: $x = \pi n$ (obvious minimum) and $x=\arccos(\frac{-1}{n-1})$, but I'm stuck here and don't know how to continue.
Any tips would be greatly appreciated! 

Remember that exponentials beat powers so your limit at fixed $x$ equals zero.
(edited)
For uniform convergence note that the function is maximized whenever $\cos x= (n+1)^{-1/2}$ and $\sin x= \{ n/(n+1)\}^{1/2}$.  The function value is then
$$\Bigl( \frac{n}{n+1} \Bigr)^{n/2} = \Bigl( 1+ \frac{1}{n+1} \Bigr)^{n/2} \to \sqrt{e} \neq 0.$$
