Question

Let $f\left(x\right)=\left\{\begin{array}{ll}x+1,& \mathrm{if}x\ge 0\\ x-1,& \mathrm{if}x<0\end{array}.\right.$Prove that $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.

Answer 1

$$\begin{gathered} f\left(x\right)=\left\{\begin{array}{ll}x+1,& x\ge 0\\ x-1,& x<0\end{array}\right. \\ \mathrm{RHL}: \\ \underset{x\to 0^{+}}{\lim }f\left(x\right) \\ =\underset{x\to 0}{\lim }\left(x+1\right) \\ \mathrm{Let}x=0+h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left(0+h+1\right) \\ =1 \\ \mathrm{LHL}: \\ \underset{x\to 0^{-}}{\lim }f\left(x\right) \\ =\underset{x\to 0^{-}}{\lim }\left(x-1\right) \\ \mathrm{Let}x=0-h,\mathrm{where}h\to 0. \\ \underset{h\to 0}{\lim }\left(0-h-1\right) \\ =-1 \\ \mathrm{LHL}\ne \mathrm{RHL} \\ \mathrm{Thus},\underset{x\to 0}{\lim }f\left(x\right)\mathrm{does}\mathrm{not}\mathrm{exist}. \end{gathered}$$