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?
can a function be differentiable at a point if the difference quotient has a finite limit but the derivative has no limit?
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derivatives
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0It is differentiable at $0$ but not continuously – 2017-01-27
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0Thats not true ! See my answer. – 2017-01-27
1 Answers
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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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0FWIW: We say that $f$ is differentiable but *not "continuously differentiable"*. – 2017-01-27
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0This was exactly my comment – 2017-01-27