# Prove that the function

Question:

Prove that the function

$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|+2 x^{2}}, & x \neq 0 \\ k & , x=0\end{array}\right.$

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

Solution:

The given function can be rewritten as:

$f(x)=\left\{\begin{array}{l}\frac{x}{x+2 x^{2}}, x>0 \\ \frac{-x}{x-2 x^{2}}, x<0 \\ k, x=0\end{array}\right.$

$\Rightarrow f(x)=\left\{\begin{array}{c}\frac{1}{2 x+1}, x>0 \\ \frac{1}{2 x-1}, x<0 \\ k, x=0\end{array}\right.$

We observe

$(\mathrm{LHL}$ at $x=0)=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0} f(-h)$

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

$(\mathrm{RHL}$ at $x=0)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)$

So, $\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$ such that $\lim _{x \rightarrow 0^{-}} f(x) \& \lim _{x \rightarrow 0^{+}} f(x)$ are independent of $k$.

Thus, $f(x)$ is discontinuous at $x=0$, regardless of the choice of $k$.