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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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    It might be worth mentioning that this is related to http://math.stackexchange.com/q/78825/78502011-11-05
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    Compute the limit of the difference quotient by hand and use the squeeze theorem.2011-11-05
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    See also http://math.stackexchange.com/questions/71613/proving-a-function-is-continuous-for-a-fixed-variable/71707#717072011-11-05
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    Another relevant thread is http://math.stackexchange.com/questions/583292011-11-05
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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 $f'(x) = c$.