Question

Let $f\left(x\right)=\{\begin{array}{cc}\underset{n\to \infty }{lim}\left(\frac{|x+1{|}^n+x^2}{|x|+x^{2n}}\right);& -6\le x<0\\ \left\{\sin x\right\};& 0\le x\le 6\end{array}$ where $\left\{k\right\}$ denotes the fractional part of $k.$
Then number of points at which $f$ is not differentiable in $(-6,6)$ is equal to

Answer 1

$f\left(x\right)=\{\begin{array}{cc}\underset{n\to \infty }{lim}\left(\frac{|x+1{|}^n+x^2}{|x|+x^{2n}}\right);& -6\le x<0\\ \left\{\sin x\right\};& 0\le x\le 6\end{array}$

When $x\in [-6,-1),$
$\underset{n\to \infty }{lim}\left(\frac{(-1{)}^n(x+1{)}^n+x^2}{-x+x^{2n}}\right)=\underset{n\to \infty }{lim}(\frac{x^n}{x^n}\times \frac{(-1{)}^n{(1+\frac{1}{x})}^n+\frac{1}{x^{n-2}}}{x^n-\frac{1}{x^{n-1}}})=0$

When $x=-1,$
$f\left(x\right)=\underset{n\to \infty }{lim}\left(\frac{|x+1{|}^n+x^2}{|x|+x^{2n}}\right)=\frac{0+1}{1+1}=\frac{1}{2}$

When $x\in (-1,0),$
$f\left(x\right)=\underset{n\to \infty }{lim}\left(\frac{(x+1{)}^n+x^2}{-x+x^{2n}}\right)=\frac{x^2}{-x}=-x$

When $x\in [0,\frac{\pi }{2}),$
$f\left(x\right)=\left\{\sin x\right\}=\sin x$
When $x=\frac{\pi }{2},$
$f\left(x\right)=\left\{\sin x\right\}=0$
When $x\in (\frac{\pi }{2},\pi ),$
$f\left(x\right)=\left\{\sin x\right\}=\sin x$
When $x=\pi ,$
$f\left(x\right)=\left\{\sin x\right\}=0$
When $x\in (\pi ,6],$
$f\left(x\right)=\left\{\sin x\right\}=\sin x-\left[\sin x\right]=\sin x-(-1)=1+\sin x$

$f\left(x\right)=\{\begin{array}{cc}0;& -6\le x<-1\\ \frac{1}{2};& x=-1\\ -x;& -1<x<0\\ \sin x;& 0\le x<\frac{\pi }{2}\\ 0;& x=\frac{\pi }{2}\\ \sin x;& \frac{\pi }{2}<x<\pi \\ 0;& x=\pi \\ 1+\sin x;& \pi <x\le 6\end{array}$

$f\left(x\right)$ is not continuous at $x=-1,\frac{\pi }{2},\pi$
And at $x=0,$ $f\left(x\right)$ is continuous but not differentiable.

Hence, $f$ is non-differentiable at $4$ points $\{-1,0,\frac{\pi }{2},\pi \}$