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

We have $f(x) = \left\{ {\begin{array}{llllllllllllllllllll}{\frac{x}{{|x| + 2{x^2}}},}&{{\rm{ if }}x \ne 0}\\{k,}&{{\rm{ if }}x = 0}\end{array}} \right.$.

At $x = 0,$ ${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} \frac{x}{{|x| + 2{x^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{(0 - h)}}{{|0 - h| + 2{{(0 - h)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h + 2{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h(1 + 2h)}} = - 1$

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{|x| + 2{x^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{0 + h}}{{|0 + h| + 2{{(0 + h)}^2}}}$

$= \mathop {\lim }\limits_{h \to 0} \frac{h}{{h + 2{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{{h(1 + 2h)}} = 1$
and $f(0) = k$

Since, ${\rm{LHL}} \ne {\rm{RHL}}$ for any value of ${\rm{k}}$.

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