Question

Prove that the function 

\[f\left( x \right) = \begin{cases}\frac{x}{\left| x \right| + 2 x^2}, & x \neq 0 \\ k , & x = 0\end{cases}\]  remains discontinuous at x = 0, regardless the choice of k.

Answer 1

The given function can be rewritten as: 

\[f\left( x \right) = \begin{cases}\frac{x}{x + 2 x^2}, x > 0 \\ \frac{- x}{x - 2 x^2}, x < 0 \\ k, x = 0\end{cases}\]

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{1}{2x + 1}, x > 0 \\ \frac{1}{2x - 1}, x < 0 \\ k, x = 0\end{cases}\]

We observe

(LHL at x = 0) =

\[\Rightarrow f\left( x \right) = \begin{cases}\frac{1}{2x + 1}, x > 0 \\ \frac{1}{2x - 1}, x < 0 \\ k, x = 0\end{cases}\]

\[\lim_{h \to 0} \frac{1}{- 2h - 1} = - 1\]

(RHL at x = 0) =

\[\lim_{h \to 0} \frac{1}{- 2h - 1} = - 1\]

\[\lim_{h \to 0} \frac{1}{2h + 1} = 1\]

So, ​

\[\lim_{h \to 0} \frac{1}{2h + 1} = 1\]

\[\lim_{x \to 0^-} f\left( x \right) \text{and} \lim_{x \to 0^+} f\left( x \right)\]

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