Question

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

Answer 1

We have, $f\left(x\right)=\{\begin{array}{cc}{x}^2-1,& x\le 1\\ -x^2-1,& x>1\end{array}$
$L.H.L.=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(x^2-1)=0$
$\Rightarrow L.H.L.=0$
$R.H.L.=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(-x^2-1)=-2$
$\Rightarrow R.H.L.=-2$
$\because L.H.L.\ne R.H.L.$
Hence $\underset{x\to 1}{lim}f\left(x\right)$ does not exist