Question

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

Answer 1

Finding limit at $x=1$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}{x}^2-1=1^2-1=0$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}{(-x}^2-1)={(-1)}^2-1=-1-1=-2$

$\therefore \underset{x\to 1^{+}}{lim}f\left(x\right)\ne \underset{x\to 1^{-}}{lim}f\left(x\right)$

$\therefore$ left hand limit and right limit are not equal

Hence $\underset{x\to 1}{lim}f\left(x\right)$ does not exist