Question

Let $f\left(x\right)=\left\{\begin{array}{ll}x+5,& \mathrm{if}x>0\\ x-4,& \mathrm{if}x<0\end{array}\right.$. Prove that $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.

Answer 1

We have,

$f\left(x\right)=\left\{\begin{array}{ll}x+5,& \mathrm{if}x>0\\ x-4,& \mathrm{if}x<0\end{array}\right.$

LHL of f(x) at x = 0

= $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }\left(0-h-4\right)=-4$

RHL of f(x) at x = 0

= $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }\left(0+h+5\right)=5$

Clearly, $\underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$

Hence, $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.