Differentiability at x = 0 via the limit definition

Since $f(0) = 0$, the difference quotient is $\frac{f(h) - f(0)}{h} = \frac{h^2 \sin(1/h) - 0}{h} = h \sin(1/h)$ for $h \neq 0$. This expression is exactly what we need to take the limit of as $h \to 0$.

Although $\sin(1/h)$ oscillates wildly and has no limit as $h \to 0$, it is always bounded between $-1$ and $1$. So $|h\sin(1/h)| \le |h|$, and by the Squeeze Theorem, $\lim_{h\to 0} h\sin(1/h) = 0$. Therefore the limit defining $f'(0)$ exists and equals $0$, so $f$ is differentiable at $x=0$ with $f'(0) = 0$, even though the graph oscillates infinitely near the origin.