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
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.
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