Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

$$\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} = \lim_{x\rightarrow a} f'(c_x) = \lim_{c_x\rightarrow a} f'(c_x)$$

Where $f:\mathbb{R}\rightarrow \mathbb{R}$, and $c_x$ is some value between $x$ and $a$ by the mean value theorem.

  1. How do you rigorously show that as $x\rightarrow a$, then $c_x\rightarrow a$ because it is always between $x$ and $a$? Something like $|x-a|<\delta \Rightarrow |c_x-a|<\delta$? I feel that just switching $c_x$ for $x$ above is kind of non-rigorous (or completely incorrect).
  2. I think my use of $c_x$ is nonstandard; I am trying to express dependence on $x$. Is there a better way to write what I have written above?

Thank you in advance!

share|cite|improve this question
up vote 1 down vote accepted
  1. Yes, it suffices to say that $c_x$ is between $a$ and $x$. Although I'm not sure what you mean by "switching $c_x$ for $x$". If you want a very detailed proof that $c_x\to a$ as $x\to a$, here goes: take some $\epsilon>0$, and then $\delta=\epsilon$; if $|x-a|<\delta$, then $|c_x-a|<\epsilon$, as you noted yourself. There, done.

    Another way, without epsilons: you always have $0\leq|c_x-a|\leq|x-a|$; then $|x-a|\to0$ as $x\to a$, and there's a standard theorem that says that in that case, $|c_x-a|\to0$ as well, which is the same as saying that $c_x\to a$.

  2. I don't see any problem with your notation. The $x$ index in $c_x$ makes it quite clear that it depends on $x$.

share|cite|improve this answer

I take it that you want to rigorously prove the first statement and you're stuck at the specified step? I would go about doing this as follows:

There exists some $L$ such that given any $\epsilon > 0$ we see that there exists some $\delta > 0$ such that $$0 < |c_x - a| < \delta \Longrightarrow |f'(c_x) - L| < \epsilon$$ However, because $a < c_x < x$ we see that $$0 < |x - a| < \delta \Longrightarrow |c_x - a| < \delta$$ Putting these two statements together, we get $$0 < |x - a| < \delta \Longrightarrow |f'(c_x) - L| < \epsilon$$ However, if by the mean value theorem, we see that $$\frac{f(x) - f(a)}{x - a} = f'(c_x) \text { for all } x$$ So we can conclude that $L = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{c_x \to a} f'(c_x)$. But $\lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$, so we see that $\lim_{c_x \to a} f'(c_x) = f'(a)$.

P.S. This is Theorem 7, Ch. 11 of M. Spivak's Calculus if you own a copy.

share|cite|improve this answer

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.