Question

Find $\underset{x\to 1}{lim}f\left(x\right),$ where $f\left(x\right)=\{\begin{array}{cc}{x}^2-1,& x\le 1\\ -x^2-1,& x>1\end{array}$

Answer 1

Given $f\left(x\right)=\{\begin{array}{cc}{x}^2-1,& x\le 1\\ -x^2-1,& x>1\end{array}$

Limit of the given function at $x=1$ exists if $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)$
$\therefore L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}{x}^2-1=(1{)}^2-1=0$
$\therefore R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(-x^2-1)=-(1{)}^2-1=-2$
Thus, $\underset{x\to 1^{+}}{lim}f\left(x\right)\ne \underset{x\to 1^{-}}{lim}f\left(x\right)$

Since, Left Hand Limit and Right-Hand Limit are not equal,
$\underset{x\to 1}{lim}f\left(x\right)$ does not exist.