Define $f(x)=x^3\cos\frac{1}{x}$ for $x\neq0$ and $f(0)=0$.
show $ f^{'} (x) $ is continuous at $x=0$ but not differentiable at $x=0$

I am studying Goldberg book and I am confused here.
the book defines : let $f$ be real valued function on an interval in real numbers. we say $f$ has a derivative at point $c$ in that interval if $$ \lim_{x \to c} \frac{f(x)-f(c)}{x-c} $$ exists and we denote it by $f^{'}(c)$.
my first question is, does this definition means that in real analysis, we do not have to check the continuity of the function to decide wether a function is differentiable at all? is this enough to form this limit and if the limit exists we say f has derivative at that point?

my second question is : is there any technique to prove the continuity of the derivative? or do I need to used $\epsilon-\delta$ definition?
I am kinda confused how to wrap up this topic.


1 Answer 1


For the first question, YES, it is enough to apply the definition and prove that the derivative exists. Continuity need not be proved separately.

For the second question you have to compute the derivative first using theorems of differentiation.

Here we get $f'(x)=3x^{2}\cos (\frac 1 x)-x\sin (\frac 1 x)$ for $x \neq 0$ and $f'(0)=0$. Now you have to check that $f'(x) \to 0$ as $x \to 0$ which is quite easy in this case since $\sin$ and $\cos$ are bounded functions.

I leave it to you to show using definition of derivative that that $f''(0)$ does not exist.

  • $\begingroup$ thank you very much for your answer $\endgroup$
    – BesMath
    Mar 28, 2020 at 23:52

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