Question

Evaluate $\underset{x\to 0}{lim}f\left(x\right),$ where
$f\left(x\right)=\{\begin{array}{cc}\frac{|x|}{x},& x\ne 0\\ 0,& x=0\end{array}$

Answer 1

Given : $\underset{x\to 0}{lim}f\left(x\right)=\{\begin{array}{cc}\frac{|x|}{x},& x\ne 0\\ 0,& x=0\end{array}$

Limit of the given function at $x=0$ exists if $\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)$

$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}f(0-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}f(-h)$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}\frac{|-h|}{-h}$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}\frac{h}{-h}$
$\Rightarrow L.H.L.=\underset{h\to 0}{lim}-1=-1$

$R.H.L.=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}f\left(h\right)$
$=\underset{h\to 0}{lim}\frac{|h|}{h}$
$=\underset{h\to 0}{lim}=1$

$L.H.L.\ne R.H.L.$.
$\therefore \underset{x\to 0}{lim}f\left(x\right)$ doesn't exist.