Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\{\begin{array}{cc}\frac{x^2-1}{x^2-2x+1};& x>1\\ \frac{1}{2};& x=1\\ \frac{x^2-1}{x^2+2x-3};& x<1\end{array}$ $\because x^2+3x-x-3$ $=x(x+3)-1(x+3)$ Lets Find LHL, ${\lim }_{x\to 1^{-}}\frac{x^2-1}{x^2-2x+1}$ and RHL, ${\lim }_{x\to 1^{+}}\frac{x^2-1}{x^2+2x-3}$ $\begin{array}{rl}& \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{(x-1)(x+1)}{(x-1{)}^2}\Rightarrow \underset{x\to 1^{+}}{\lim }\frac{(x-1)(x+1)}{(x+3)(x-1)}\\ & \Rightarrow \underset{x\to 1^{-}}{\lim }\frac{x+1}{x-1}=\infty \Rightarrow \underset{x\to 1^{+}}{\lim }\frac{x+1}{x+3}=\frac{2}{4}=\frac{1}{2}\end{array}$ $\therefore$ LHL $\ne f(1)\Rightarrow f(x)$ is not continuous at $x=1$