# Find the value

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.