Answer
The graph of $g(x)$ does not have a tangent at the origin.
Work Step by Step
We have $g'(0)=\lim\limits_{h\to0}\dfrac{g(h+0)-g(0)}{h}=\lim\limits_{h\to0}\dfrac{g(h)-g(0)}{h}$
As we know that when $h\to0$, $h\ne 0$ , thus $g(h)=h\sin(\dfrac{1}{h})$
and $g'(0)=\lim\limits_{h\to0}\dfrac{h\sin(\dfrac{1}{h})-0}{h}=\lim\limits_{h\to0}\dfrac{h\sin(\dfrac{1}{h})}{h}$
$\implies g'(0)=\lim\limits_{h\to0}\sin(\dfrac{1}{h})$
As $h\to0$, then $\lim\limits_{h\to0} \sin(\dfrac{1}{h})$ does not exist, because the graph of $\sin(\dfrac{1}{h})$ will be oscillating at high peak to get a definite value.
Hence, the graph of $g(x)$ does not have a tangent at the origin.
