Question

Prove that the function
$f\left(x\right)=\left\{\begin{array}{ll}\frac{x}{\left|x\right|+2x^2},& x\ne 0\\ k,& x=0\end{array}\right.$
remains discontinuous at x = 0, regardless the choice of k.

Answer 1

The given function can be rewritten as:
$f\left(x\right)=\left\{\begin{array}{c}\frac{x}{x+2x^2},x>0\\ \frac{-x}{x-2x^2},x<0\\ k,x=0\end{array}\right.$
$\Rightarrow f\left(x\right)=\left\{\begin{array}{c}\frac{1}{2x+1},x>0\\ \frac{1}{2x-1},x<0\\ k,x=0\end{array}\right.$

We observe
(LHL at x = 0) = $\underset{x\to 0^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0-h\right)=\underset{h\to 0}{\lim }f\left(-h\right)$
=$\underset{h\to 0}{\lim }\frac{1}{-2h-1}=-1$

(RHL at x = 0) = $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(0+h\right)=\underset{h\to 0}{\lim }f\left(h\right)$
=$\underset{h\to 0}{\lim }\frac{1}{2h+1}=1$


So, ​$\underset{x\to 0^{-}}{\lim }f\left(x\right)\ne \underset{x\to 0^{+}}{\lim }f\left(x\right)$ such that $\underset{x\to 0^{-}}{\lim }f\left(x\right)\&\underset{x\to 0^{+}}{\lim }f\left(x\right)$ are independent of k.

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