number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$ 
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
   and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
Then number of points where $g(x)$ is not differentiable.

$\bf{My\; Try::}$ We can write $$f(x) =\lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}= \left\{\begin{matrix}
 \frac{2}{x+1}\;\;,&x+1<-1\Rightarrow x<-2 \\ 
\frac{x^2}{x^2+1}\;\;, & -1<x+1<1\Rightarrow -2<x<0\\ 
 \frac{2}{x+1}\;\;, & x+1>1\rightarrow x>0 \\ 
1\;\;, & x+1=1\Rightarrow x=0\\ 
 \frac{3}{2}\;\;,& x+1=-1\Rightarrow x=-2 
\end{matrix}\right.$$
Now How can i solve after that help Required, Thanks

if you plot function f, you will see $-2 \lt f \lt 2$ with 2 discontinuity points at -2, 0, which are not differentiable.
Next you can say the following x are not differentiable, $$\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right) = \pm n\pi$$
This leads to the points satifying the following equation.
$$\frac{2f(x)}{1+(f(x))^2} = 0$$ or $$f(x) = 0$$ 
There is only one, i.e. x=0. 
So the answer is two non differentiable points -2 and 0.

