# Prove that the function f defined by

Question:

Prove that the function defined by

$f(x)= \begin{cases}\frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k, & x=0\end{cases}$

remains discontinuous at = 0, regardless the choice of k.

Solution:

Finding the left hand and right hand limit for the given function, we have

$\lim _{x \rightarrow 0^{+}} f(x)=\frac{x}{|x|+2 x^{2}}=\lim _{h \rightarrow 0} \frac{0-h}{|0-h|+2(0-h)^{2}}$

$=\lim _{h \rightarrow 0} \frac{-h}{h+2 h^{2}}=\lim _{h \rightarrow 0} \frac{-h}{h(1+2 h)}$

$=\lim _{h \rightarrow 0} \frac{-1}{1+2 h}=\frac{-1}{1+2(0)}=-1$

$\lim _{x \rightarrow 0^{-}} f(x)=\frac{x}{|x|+2 x^{2}}=\lim _{h \rightarrow 0} \frac{0+h}{|0+h|+2(0+h)^{2}}$

$=\lim _{h \rightarrow 0} \frac{h}{h+2 h^{2}}=\lim _{h \rightarrow 0} \frac{h}{h(1+2 h)}=\frac{1}{1+0}=1$

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

Now, as the left hand limit and the right hand limit are not equal and the value of both the limits are a constant.

Hence, regardless the choice of k, the given function remains discontinuous at x = 0.