Answer
The graph of $f$ has a tangent at the origin.
Work Step by Step
$f(x)=x^2\sin(1/x)$ for $x\ne0$ and $f(x)=0$ for $x=0$
For the graph of $f$ to have a tangent at the origin, the value of its slope there, the derivative $f'(0)$, must exist.
In other words, we will see if the formula for $f'(0)$ exists or not. In detail,
$$f'(0)=\lim_{h\to0}\frac{f(h+0)-f(0)}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$
Remember here that
- For $h\to0$, $h$ does not equal $0$, so $f(h)=h^2\sin(1/h)$
- For $f(0)$, having $x=0$ means $f(0)=0$
$$f'(0)=\lim_{h\to0}\frac{h^2\sin(1/h)-0}{h}=\lim_{h\to0}\frac{h^2\sin(1/h)}{h}$$
$$f'(0)=\lim_{h\to0}\Big(h\sin(1/h)\Big)$$
It is known that $-1\le\sin(1/h)\le1$
So, $-h\le h\sin(1/h)\le h$
Because $\lim_{h\to0}h=\lim_{h\to0}-h=0$, according to the Sandwich Theorem, $$f'(0)=\lim_{h\to0}\Big(h\sin(1/h)\Big)=0$$
In other words, $f'(0)$ does exist, and so does the slope of the tangent to $f$ at the origin. Therefore, the graph of $f$ has a tangent at the origin.
