Question

$Find{lim}_{x\to 1}f\left(x\right),wheref\left(x\right)=\{x^2-1,x\le 1{-x}^2-1,x>1\}$

Answer 1

$f\left(x\right)=\{\begin{array}{c}{x}^2-1,x\le 1\\ {-x}^2-1,x>0\end{array}$

${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{-}}(x^2-1)=1^2-1=0{lim}_{x\to 1^{+}}f\left(x\right)={lim}_{x\to 1^{+}}({-x}^2-1)=-1-1=-2As{lim}_{x\to 1^{-}}f\left(x\right)\ne {lim}_{x\to 1^{+}}f\left(x\right)$

So limit of this function does not exist