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
 A: Hint: using the formula $\tan ^{ 2 }{ \frac { x }{ 2 } = } \frac { 1-\cos { x }  }{ \sin { x }  } $  we can transform $g(x)$ function so :$$g\left( x \right) =\tan  \left( \frac { 1 }{ 2 } \arcsin  \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } }  \right)  \right) =\\ =\pm \sqrt { \frac { 1-\cos { \left( \arcsin  \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } }  \right)  \right)  }  }{ \sin { \left( \arcsin  \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } }  \right)  \right)  }  }  } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1-{ \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } }  \right)  }^{ 2 } }  }{ \left( \frac { 2f(x) }{ 1+(f(x))^{ 2 } }  \right)  }  } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1+1+2f^{ 2 }\left( x \right) +f^{ 4 }\left( x \right) -4f^{ 2 }\left( x \right)  }  }{ 2f\left( x \right)  }  } =\\ =\pm \sqrt { \frac { 1-\sqrt { 1+{ \left( 1-f^{ 2 }\left( x \right)  \right)  }^{ 2 } }  }{ 2f\left( x \right)  }  } $$
hope it will help
A: First note that, for every $y$, $2y/(1+y^2)\in[-1,1]$.
Consider $g(y)=\frac{1}{2}\arcsin(2y/(1+y^2))$, so
$$
g'(y)=\frac{1}{\sqrt{1-\dfrac{4y^2}{(1+y^2)^2}}}
\cdot\frac{1+y^2-2y^2}{(1+y^2)^2}=\frac{2}{1+y^2}\frac{1-y^2}{|1-y^2|}
$$
Therefore $g'(y)$ does not exist for $y=\pm1$ and
$$
g'(y)=\begin{cases}
\dfrac{1}{1+y^2} & \text{if $|y|<1$}
\\[4px]
-\dfrac{1}{1+y^2} & \text{if $|y|<1$}
\end{cases}
$$
Thus
$$
g(y)=\begin{cases}
a+\arctan y & \text{if $|y|<1$}
\\[4px]
b-\arctan y & \text{if $|y|<1$}
\\[4px]
-\pi/4 & \text{if $y=-1$}
\\[4px]
\pi/4 & \text{if $y=1$}
\end{cases}
$$
Since $g(0)=0$, we have $a=0$; since $\lim_{y\to\infty}g(y)=0$, we have $b=\pi/2$.
Thus
$$
\tan\left(\frac{1}{2}\arcsin\frac{2y}{1+y^2}\right)=\tan(g(y))=
\begin{cases}
y & \text{if $|y|<1$}
\\[4px]
\dfrac{1}{y} & \text{if $|y|>1$}
\\[4px]
-1 & \text{if $y=-1$}
\\[4px]
1 & \text{if $y=1$}
\end{cases}
$$
A: 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.

