Find the value

Question:
Let $f(x)= \begin{cases}\frac{x-|x|}{x}, & x \neq 0 \\ 2, & x=0\end{cases}$

Show that $\lim _{x \rightarrow 0} f(x)$ does not exist

Solution:

Left Hand Limit(L.H.L.):

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x-|x|}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{x-(-x)}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{x+x}{x}$

$=\lim _{x \rightarrow 0^{-}} \frac{2 x}{x}$

$=\lim _{x \rightarrow 0^{-}} 2$

$=2$

Right Hand Limit(R.H.L.):

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{x-|x|}{x}$

$=\lim _{x \rightarrow 0^{+}} \frac{x-(x)}{x}$

$=\lim _{x \rightarrow 0^{+}} \frac{0}{x}$

$=\lim _{x \rightarrow 0^{+}} 0$

$=0$

$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$

Thus, $\lim _{x \rightarrow 0} f(x)$ does not exist.