Question

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

Answer 1

$LHL={lim}_{x\to 0}f\left(x\right)$

$={lim}_{x\to 0^{-}}(x-4)$ $\left[\begin{array}{c}\text{as}x\to 0^{-}\\ \Rightarrow x<0\text{slightly}\end{array}\right]$

=(0-4)

=-4

$RHL={lim}_{x\to 0^{+}}f\left(x\right)$

$={lim}_{x\to 0^{+}}(x+5)$ $\left[\begin{array}{c}\text{as}x\to 0^{+}\\ \Rightarrow x<0\text{slightly}\end{array}\right]$

=(0+5)

=5

Clearly, $li{m}_{x\to 0^{-}}f\left(x\right)\ne li{m}_{x\to 0^{+}}f\left(x\right)$

Hence $li{m}_{x\to 0}f\left(x\right)$ does not exist.