1
$\begingroup$

My friend asks me about this problem.

$f$ is differentiable on $(a,b)$, for $c\in (a,b)$, exists a sequence $x_n\to c$ and \lim f'(x_n)=f'(c), how to show that f' doesn't have to be continuous at $c$?

I know 'there exists' such a sequence does not guarantee that it's true for any sequence converging to $c$, but I can't think of a counterexample. Can anybody prove it or find a counterexample? This problem is kinda interesting and I want to know the answer:) Thanks!!!

  • 1
    Trivial counterexample : take $f$ such that $f'$ is not continuous at $0$ and take $x_n=0$.2011-11-17

1 Answers 1

7

Here's a "hand waving" proof that an example exists:

Take $f(x)=\cases{x^2\sin (\textstyle{1\over x}), &x\ne 0\cr 0, &x=0}$

The graph of this function is

enter image description here

$f$ is differentiable for all $x\ne0$. In fact, for $x\ne0$: $ f'(x) = 2x\sin (\textstyle{1\over x})- \cos(\textstyle{1\over x}). $

One can show $f'(0)$ exists and is equal to $0$.

Indeed, for $x\ne0$: $ \eqalign{ {f(x)-f(0)\over x-0} = x\sin(\textstyle{1\over x})\ \buildrel{x\rightarrow 0}\over \longrightarrow\ 0. } $

Also, one can find a sequence $x_n\searrow 0$ for which $f'(x_n)=0$ for each $n$ (look at the horizontal tangents of the graph). But $f'$ is not continuous at zero since it takes values of both 0 and $a$ in every neighborhood of $0$ for some $a>0$ (to see this, look at the formula for $f'(x)$ above and consider: $2x\sin(\textstyle{1\over x})$ becomes small for small $x$ and $-\cos(\textstyle{1\over x})$ takes the value 1 at some point of every nhood of $0$).