Find limit of $f_n(x)= (\cos(x))(\sin(x))^n \sqrt{n+1}$ $f_n(x)$ on $ \Bbb{R}$ defined by 
$$f_n(x)= (\cos(x))(\sin(x))^n  \sqrt{n+1}$$ Then
 Is It  converges uniformly ?
I think first we must find limit of $f_n$ , I find limit for 0 and $\frac{\pi}{2}$ ,but I can't find for every point.
 A: The limit is $0$. If $x=2k\pi$, $(k\in \mathbb{Z})$, it is clear. For $x\not=2k\pi$, change $n$ to a continuous variable $y$ so that you can use the L'Hôpital's rule, and prove that $\lim_{y\rightarrow+\infty}\vert \sin x\vert^{y}\sqrt{y+1}=0.$ You will have 
$$\lim_{y\rightarrow+\infty}\dfrac{\sqrt{y+1}}{\frac{1}{\vert \sin x\vert^{y}}}=\frac{\infty}{\infty},$$
and L'Hôpital's can be used. Got it?
A: A... If $\sin x=0$ or $\cos x=0$ then $f_n(x)=0$ for every $n.$ If $0<|\sin x|<1$ let $|\sin x|=1/(1+y) .$ Since $y>0$ we have $(1+y)^n\geq 1+n y, $ so $0<|f_n(x)|<|\sin x|^n\sqrt {1+n}<(1+n y)^{-1}\sqrt {1+n}.$ So $f_n(x)\to 0$ as $n\to \infty.$
B... Let $g_n(x)=\cos x \sin^n x.$ For any $x$ there exists $x'\in [0,\pi /2]$ with $|g_n(x')|=|g_n(x)|$.
C...  We have $g'_n(x)=(-\sin^2 x+n\cos^2 x)\sin^{n-1} x .$ Now $g'_n(0)\geq 0$ for $x\in [0,\arctan \sqrt n],$ while $g'(n)<0$ for $x\in (\arctan \sqrt n,\pi /2].$ Therefore $\max_{x\in [0,\pi /2]}g_n(x)=g_n(\arctan \sqrt n)$ and $\min_{x\in [0,\pi /2]}g_n(x)=\min (g_n(0),g_n(\pi /2))=0.$
D...  From B. and C. we have $\max |g_n(x)|= g_n(\arctan \sqrt n).$
E...  For brevity let $x_n=\arctan \sqrt n.$ We  have $f_n(x_n)=(1+1/n)^{-n/2}$ which tends to $1/\sqrt e$ as $n\to \infty$, so $f_n$ does not converge uniformly to $0.$
