Answer
The graph of $g$ does not have a tangent at the origin.
Work Step by Step
$g(x)=x\sin(1/x)$ for $x\ne0$ and $g(x)=0$ for $x=0$
For the graph of $g$ to have a tangent at the origin, the value of its slope there, the derivative $g'(0)$, must exist.
In other words, we will see if the formula for $g'(0)$ exists or not. In detail,
$$g'(0)=\lim_{h\to0}\frac{g(h+0)-g(0)}{h}=\lim_{h\to0}\frac{g(h)-g(0)}{h}$$
Remember here that
- For $h\to0$, $h$ does not equal $0$, so $g(h)=h\sin(1/h)$
- For $g(0)$, having $x=0$ means $g(0)=0$
$$g'(0)=\lim_{h\to0}\frac{h\sin(1/h)-0}{h}=\lim_{h\to0}\frac{h\sin(1/h)}{h}$$
$$g'(0)=\lim_{h\to0}\sin(1/h)$$
As $h\to0$, $\lim_{h\to0}\sin(1/h)$ does not exist, because the graph of $\sin(1/h)$ oscillates too much to reach for a definite value.
In other words, $g'(0)$ does not exist, and neither does the slope of the tangent to $g$ at the origin. Therefore, the graph of $g$ does not have a tangent at the origin.