2
$\begingroup$

Let $f:[a,b]\to \mathbb{R}$ be differentiable on $(a,b)$ with derivative $g=f^{\prime}$ there.

Assertion: If $\lim_{x\to b^{-}}g(x)$ exists and is a real number $\ell$ then $f$ is differentiable at $b$ and $f^{\prime}(b)=\ell$?

Is this assertion correct? If so provide hints for a formal $\epsilon-\delta$ argument. If not, can it be made true if we strengthen some conditions on $g$ (continuity in $(a,b)$ etc.)? Provide counter-examples.

I personally think that addition of the continuity of $g$ in the hypothesis won't change anything as for example $x\sin \frac{1}{x}$ has a continuous derivative in $(0,1)$ but its derivative oscillates near $0$. I also know that the converse of this is not true.

Also if that limit is infinite, then $f$ is not differentiable at $b$ right?

  • 0
    Then write $\left.f'\right|_{(a,b)}$2012-11-19

2 Answers 2

3

Since $ f(b+h)-f(b)=f'(\xi)h $ for some $\xi \in (b-h,h)$, you can let $h \to 0^{-}$ and conclude that $f'(b)=\lim_{x \to b-}f'(x)$. On the other hand, consider $f(x)=x^2 \sin \frac{1}{x}$. It is easy to check that $\lim_{x \to 0} f'(x)$ does not exist, and yet $f'(0)=0$.

Edit: this answer assumes tacitly that $f$ is continuous at $b$. The question does not contain this assumption, although I believe that it should be clear that a discontinuous function can't be differentiable at all.

  • 0
    In this case, $f$ must be discontinuous at $b$, so the necessary condition for differentiability is violated.2012-11-20
1

This is clearly false as stated. For example, consider $f:[0,1] \to \mathbb{R}$, $f(1)=1, f(x)=0$ otherwise.

  • 0
    @no-0-one-1 I assumed (see my remark) that $f$ is continuous at $b$. It is meaningless to prove differentiability at points where the function is discontinuous. The answer by Chris Eagle is correct, but I suspect that you were thinking of a continuous function, since otherwise the answer would be trivial.2012-11-20