$h(x)=\begin{cases}
x^2\sin(1/x)& \text{if}\hspace{.2cm} x \neq 0\\
0,& \text{otherwise}
\end{cases}$. What is $h'(0)$?
Answer & explanation
Correct answer: option 3
$h'(0)=\lim_{x \to 0}\frac{h(x)-h(0)}{x-0}=\lim_{x \to 0}x\sin{1/x}=0$
Correct answer: option 3