Problem. Let $\displaystyle g(x)=x^{2}sin( \frac {1}{x} )$, find g'(0) and prove that g'(x) is not continuous at x=0.

My work:

$\displaystyle g'(0) = \lim _{x \rightarrow 0 } \frac {g(x)-g(0)}{x-0} = \lim _{x \rightarrow 0 } \frac {x^{2}sin( \frac {1}{x} )-g(0)}{x}$, but g(0) is undefined, so how do you find the derivative?

Thanks.