$f\left(x\right) = x^5 \text{sgn} \; x$, where $\text{sgn} \; x = \begin{cases} \dfrac{\left|x\right|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ $\;\;$ ($\text{sgn} \; x$ is the signum function) Examine the continuity and differentiability of $f\left(x\right)$ at $x=0$.
$\begin{aligned}
\left|x\right| = \begin{cases}
x, & x > 0 \\
-x, & x < 0
\end{cases}
\end{aligned}$
$\begin{aligned}
\therefore \; \text{sgn} \; x = \begin{cases}
\dfrac{x}{x}=1, & x > 0 \\
& \\
-\dfrac{x}{x}=-1, & x < 0 \\
& \\
0, & x=0
\end{cases}
\end{aligned}$
$\begin{aligned}
\therefore \; f\left(x\right) = x^5 \text{sgn} \; x = \begin{cases}
x^5, & x > 0 \\
-x^5, & x < 0 \\
0, & x=0
\end{cases}
\end{aligned}$
$\begin{aligned}
\text{Left Hand Derivative } \left(\text{at }x=0\right) = Lf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0-h\right)-f\left(0\right)}{-h} \\
& = \lim\limits_{h \to 0} \; \dfrac{-\left(-h\right)^5 - 0}{-h} \\
& = \lim\limits_{h \to 0} \; \left(-h^4\right) = 0
\end{aligned}$
$\begin{aligned}
\text{Right Hand Derivative } \left(\text{at }x=0\right) = Rf'\left(0\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(0+h\right)-f\left(0\right)}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{h^5 - 0}{h} \\
& = \lim\limits_{h \to 0} \; \left(h^4\right) = 0
\end{aligned}$
$\therefore$ $\;$ $Lf'\left(0\right) = Rf'\left(0\right) = 0$
$\implies$ $f\left(x\right)$ is differentiable at $x=0$
$\because$ $\;$ $f\left(x\right)$ is differentiable at $x=0$, $f\left(x\right)$ is also continuous at $x=0$.