Question:
Write the derivative of $f(x)=|x|^{3}$ at $x=0$.
Solution:
Given: $f(x)=|x|^{3}= \begin{cases}x^{3}, & x \geq 0 \\ -x^{3}, & x<0\end{cases}$
$(\mathrm{LHD}$ at $x=0)$
$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{x}$
$=\lim _{h \rightarrow 0} \frac{h^{3}}{-h}$
$=0$
(RHD at x = 0)
$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{x}$
$=\lim _{h \rightarrow 0} \frac{h^{3}-0}{h}$
$=0$
and $f(0)=0$
Thus, $(\mathrm{LHD}$ at $x=0)=(\mathrm{RHD}$ at $x=0)=f(0)$
Hence, $\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0}=f^{\prime}(0)=0$