Question:
Let $f(x)= \begin{cases}\frac{3 x}{|x|+2 x} & , x \neq 0 \\ 0, & x=0\end{cases}$
Show that $\lim _{x \rightarrow 0} f(x)$ does not exist.
Solution:
Left Hand Limit(L.H.L.):
$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{3 x}{|x|+2 x}$
$=\lim _{x \rightarrow 0^{-}} \frac{3 x}{(-x)+2 x}$
$=\lim _{x \rightarrow 0^{-}} \frac{3 x}{x}$
$=\lim _{x \rightarrow 0^{-}} 3$
$=3$
Right Hand Limit(R.H.L.):
$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{3 x}{|x|+2 x}$
$=\lim _{x \rightarrow 0^{+}} \frac{3 x}{(x)+2 x}$
$=\lim _{x \rightarrow 0^{+}} \frac{3 x}{3 x}$
$=\lim _{x \rightarrow 0^{+}} 1$
= 1
Since
$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$
Thus, $\lim _{x \rightarrow 0} f(x)$ does not exist.