Question

Show that

\[f\left( x \right)\] = \begin{cases}\frac{x - \left| x \right|}{2}, when & x \neq 0 \\ 2 , when & x = 0\end{cases}

is discontinuous at x = 0.

 

Answer 1

The given function can be rewritten as:

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

\[\Rightarrow\]

\[f\left( x \right) = \begin{cases}0, when x > 0 \\ x, when x < 0 \\ 2, when x = 0\end{cases}\]

We observe
(LHL at x = 0) = \[\lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right) = \lim_{h \to 0} f\left( - h \right)\]

\[= \lim_{h \to 0} \left( - h \right) = 0\]

(RHL at x = 0) = 

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

\[\lim_{h \to 0} 0 = 0\]

And, 

\[f\left( 0 \right) = 2\]

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

Thus, f(x) is discontinuous at x = 0.