if x+ 1 f(x) = { x* a = 1 1 if x= 1

The function is given ,

$f\left(x\right)=\left(\begin{array}{c}\frac{x^2-x}{x^2-1} , if x\cancel{\ne }1\\ 1, if x=1\end{array}\right)$ for $a=1.$

To check the function discontinuity,

The left hand limit is,

$\begin{array}{rcl}\underset{h\to 0}{\lim }f\left(1-h\right)& =& \underset{h\to 0}{\lim }\frac{{\left(1-h\right)}^2-\left(1-h\right)}{{\left(1-h\right)}^2-1}\\ & =& \underset{h\to 0}{\lim }\left[\frac{1+h^2-2h-1+h}{1+h^2-2h-1}\right]\\ & =& \underset{h\to 0}{\lim }\left[\frac{h^2-h}{h^2-2h}\right]\\ & =& \underset{h\to 0}{\lim }\left[\frac{h-1}{h-2}\right]\\ & =& \frac{1}{2}\end{array}$

The right hand limit is, 

$\begin{array}{rcl}\underset{h\to 0}{\lim }f\left(1+h\right)& =& \underset{h\to 0}{\lim }\frac{{\left(1+h\right)}^2-\left(1+h\right)}{{\left(1+h\right)}^2-1}\\ & =& \underset{h\to 0}{\lim }\left[\frac{1+h^2+2h-1-h}{1+h^2+2h-1}\right]\\ & =& \underset{h\to 0}{\lim }\left[\frac{h^2+h}{h^2+2h}\right]\\ & =& \underset{h\to 0}{\lim }\left[\frac{h+1}{h+2}\right]\\ & =& \frac{1}{2}\end{array}$

The function at $a=1$ ,is $f\left(1\right)=1.$

Thus , the LHL$=$RHL$\cong f\left(1\right)$.