Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm stuck on this old qualifier problem. I suppose one could do it using the basic definitions of continuity and differentiability, but is there a simpler way? (For example, using DCT, FTC, Lebesgue differentiation theorem, etc.)

Let $f:\mathbb{R} \mapsto \mathbb{R}$ be continuous. Suppose $f$ is differentiable away from $0$ and lim$_{x \to 0} f^\prime(x)$ exists. Show $f^\prime(0)$ exists.

share|cite|improve this question
up vote 6 down vote accepted

By the mean value theorem, there is a $c_x\in (0,x)$ resp. $c_x\in (x,0)$, depending on whether $x > 0$ or $x < 0$, such that

$$\frac{f(x)-f(0)}{x} = f'(c_x).$$

As $x\to 0$, by the squeeze lemma, also $c_x\to 0$, hence

$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = \lim_{x\to 0}f'(c_x)$$


share|cite|improve this answer
Nice answer! Thanks. – StrangerLoop Jul 5 '14 at 14:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.