Does the graph of
$$
g(x)=\left\{\begin{array}{ll}{x \sin (1 / x),} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.
$$
have a tangent at the origin? Give reasons for your answer.
Vertical Tangents We say that a continuous curve $y=f(x)$ has a vertical tangent at the
point where $x=x_{0}$ if $\lim _{k \rightarrow 0}\left(f\left(x_{0}+h\right)-f\left(x_{0}\right)\right) / h=\infty$ or $-\infty$ . For example, $y=x^{1 / 3}$ has a vertical tangent at $x=0$ (see accompanying figure):
$$
\begin{aligned} \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} &=\lim _{h \rightarrow 0} \frac{h^{1 / 3}-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{h^{2 / 3}}=\infty \end{aligned}
$$
However, $y=x^{2 / 3}$ has $n o$ vertical tangent at $x=0$ (see next figure):
$$
\begin{aligned} \lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h} &=\lim _{h \rightarrow 0} \frac{h^{2 / 3}-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{h^{1 / 3}} \end{aligned}
$$
does not exist, because the limit is $\infty$ from the right and $-\infty$ from the left.