$h(x)=\begin{cases} x^2\sin(1/x)& \text{if}\hspace{.2cm} x \neq 0\\ 0,& \text{otherwise} \end{cases}$. What is $h'(0)$?

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$h'(0)=\lim_{x \to 0}\frac{h(x)-h(0)}{x-0}=\lim_{x \to 0}x\sin{1/x}=0$