Q. $f(x) = \left\{ \begin{array}{r} 2x,\ \ \ \ \ \ \ \ \ \ x < 0 \\ 2x + 1,\ \ x \geq 0 \end{array} \right.\$. Then
A) $f(x)$ is continuous at $x = 0$
B) $f(|x|)$ is continuous at $x = 0$
C) $f(x)$ is discontinuous at $x = 0$
D) None of the above
Explanation
(c)
$\lim_{x \rightarrow 0^{-}}{f(x)} = \lim_{h \rightarrow 0}2(0 - h) = 0$
And $\lim_{x \rightarrow 0^{+}}{f(x)} = \lim_{h \rightarrow 0}\ 2(0 + h) + 1 = 1$
$\because\ \ \ \lim_{x \rightarrow 0^{-}}{f(x)} \neq \lim_{x \rightarrow 1^{+}}{f(x)}\$
$\therefore\ \ \ f(x)$ is discontinuous at $x = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.