Let f(x) = x²sin1/x for x ≠ 0, and f(0) = 0…

Step 1. The derivative at $0$. $\dfrac{f(h) - f(0)}{h} = h\sin\dfrac{1}{h}$. Its size is at most $|h|$, which tends to $0$. So $f'(0) = 0$.

Step 2. The derivative elsewhere. For $x \ne 0$, $f'(x) = 2x\sin\dfrac{1}{x} - \cos\dfrac{1}{x}$.

Step 3. Two sample points. At $x = \dfrac{1}{2\pi}$ we have $\sin 2\pi = 0$ and $\cos 2\pi = 1$, so $f'(x) = -1$. At $x = \dfrac{1}{\pi}$ we have $\sin\pi = 0$ and $\cos\pi = -1$, so $f'(x) = 1$.

Step 4. The limit. Taking $x = \dfrac{1}{2\pi n}$ and $x = \dfrac{1}{(2n + 1)\pi}$ with large $n$ gives points as close to $0$ as we like, where $f'$ is $-1$ and $1$. So $f'$ has no limit at $0$.