Limit and mean value theorem $$\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} = \lim_{x\rightarrow a} f'(c_x) = \lim_{c_x\rightarrow a} f'(c_x)$$
Where $f:\mathbb{R}\rightarrow \mathbb{R}$, and $c_x$ is some value between $x$ and $a$ by the mean value theorem.


*

*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).

*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!
 A: 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. 
A: *

*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 $c_x\to a$.


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