Question

If $f(x)=|x|+|\sin x|$ for $x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, then its left hand derivative at $x=0$ is

• A. 0
• B. -1
• C. -2
• D. -3

Answer 1

Answer: (C)

$f(x)=|x|+|\sin x|$
$LHD =\displaystyle\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{0-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{|0-h|+|\sin (0-h)|-(0+0)}{0-h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{h+\sin h}{-h}=-\displaystyle\lim _{h \rightarrow 0}\left(1+\frac{\sin h}{h}\right)$
$=-(1+1)=-2$