Answer
The graph of $f(x)$ has a tangent at the origin.
Work Step by Step
$f'(0)=\lim_{h\to0}\frac{f(h+0)-f(0)}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}$
As we know that $h\to 0$, $h$ does not equal $0$, so $f(h)=h^2\sin(\dfrac{1}{h})$ and for $f(0)$ at $x=0 \implies f(0)=0$
Now, $f'(0)=\lim\limits_{h\to 0}\dfrac{h^2\sin(1/h)-0}{h}=\lim\limits_{h\to 0} \dfrac{h^2\sin(1/h)}{h}=\lim\limits_{h\to 0}(h\sin(1/h)\Big)$
Also, $-1\le\sin(1/h)\le1$
This implies that , $-h\leq h\sin(\dfrac{1}{h})\leq h$
As per the Sandwich Theorem, we have $f'(0)=\lim\limits_{h\to 0}(h\sin(\dfrac{1}{h}))=0$
This means that the slope of the tangent to $f(x)$ is at the origin. and the graph of $f(x)$ has a tangent at the origin.
