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Let $f(x)=x^2\sin(1/x)$ for $x≠ 0$ and $f(0)=0$ for $x=0$.

Is f' continuous at $0$?

My attempt: f'(x)=2x\sin(1/x)-\cos(1/x). Since when $x$ goes to $0$, the limit of $\cos(1/x)$ does not exist, it is not continuous. But I'm not sure since we did define $f(0)=0$...

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    @mixedmath hmm the answers posted below seem to differ - just wanted to let you know :)2011-11-06

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Your argument is correct: for $x\ne 0$ the derivative is f\;'(x)=2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)\;, and \lim_{x\to 0}\;f\;'(x) = \lim_{x\to 0} \left(2x\sin\left(\frac1x\right)-\cos\left(\frac1x\right)\right)=-\lim_{x\to 0}\;\cos\left(\frac1x\right) does not exist, so it cannot equal f\;'(0). The latter of course, is $\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\;h\sin\left(\frac1h\right)=0\;,$ but you don’t need that to determine that f\;' is not continuous at $0$.

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You are correct: f' is not continuous at $0$. Nevertheless, $f$ is differentiable at $0$, with f'(0) = 0. This doesn't come from formally differentiating the expression for $f$, but by working directly from the definition of differentiability.

Incidentally, although the derivative of a differentiable function doesn't have to be continuous, it does have to satisfy the intermediate value property: for every $a$ and $b$, if $c$ lies between f'(a) and f'(b) then there exists $x$ between $a$ and $b$ such that f'(x) = c.