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In the case of x^2*sin((5x+4)/x). the difference quotient has limit 0 for x->0, but the derivative itself is 2x*sin((5x+4)/x)+ 4*cos((5x+4)/x), which has no limit for x->0. So is the function differentiable at 0?

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    It is differentiable at $0$ but not continuously2017-01-27
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    Thats not true ! See my answer.2017-01-27

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I suppose that you have the following function:

$f(x)=x^2*\sin((5x+4)/x)$ for $x \ne 0$ and $f(0)=0$

Then

$| \frac{f(x)-f(0)}{x-0}|=|x|*|\sin((5x+4)/x)| \le |x|$,

therefore

$\frac{f(x)-f(0)}{x-0} \to 0$ for $x \to 0$.

This shows that $f$ is differentiable at $x=0$.

Furthermore the limit $\lim_{x \to 0}f'(x)$ does not exist. This means that $f'$ is not continuous at $x=0$.

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    FWIW: We say that $f$ is differentiable but *not "continuously differentiable"*.2017-01-27
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    This was exactly my comment2017-01-27