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This is my second question for this proof. The first question can be found here.

My textbook states the proof as follows:

Let $f$ be a continuous one-to-one function defined on an interval, and suppose that $f$ is differentiable at $f^{-1}(b)$, with derivative $f'(f^{-1}(b)) \not = 0$. Then $f^{-1}$ is differentiable at $b$, and

$(f^{-1})'(b) = \dfrac{1}{f'(f^{-1}(b))}$.

Let $b = f(a)$. Then $\lim_{h \to 0} \dfrac{f^{-1}(b + h) - f^{-1}(b)}{h} = \lim_{h \to 0} \dfrac{f^{-1}(b + h) - a}{h} $

Now every number $b + h$ in the domain of $f^{-1}$ can be written in the form $b + h = f(a + k)$ for a unique $k$ (we should write $k(h)$, but we will stick with $k$ for simplicity). Then

$\lim_{h \to 0} \dfrac{f^{-1}(b + h) - a}{h} $

$= \lim_{h \to 0} \dfrac{f^{-1}(f(a + k)) - a}{f(a + k) - b} $

$= \lim_{h \to 0} \dfrac{k}{f(a + k) - f(a)} $

We are clearly on the right track! It is not hard to get an explicit expression for $k$; since

$b + h = f(a + k)$

we have

$k = f^{-1}(b + h) - f^{-1}(b)$

The function $f^{-1}$ is continuous at $b$. This means that $k$ approaches $0$ as $h$ approaches $0$. Since

$\lim_{k \to 0} \dfrac{f(a + k)) - f(a)}{k} $

and the proof continues ...

I don't understand how the proof went from

$\lim_{h \to 0} \dfrac{k}{f(a + k) - f(a)} $

to

$\lim_{k \to 0} \dfrac{f(a + k)) - f(a)}{k} $

I have unsuccessfully tried to algebraically manipulate the first expression to get the second one. Also, notice the change of the limit variable from $h \to 0$ to $k \to 0$ -- this is another change that I can't reason about.

I would greatly appreciate it if people could please take the time to explain this step.

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    They not "switch" from one limit to another, simply they need to use the fact that $f$ is differentiable, so is, that limit exists :) I bet the proof continues with $\lim_{k \to 0}=1/(f(a+k)-f(a)/k)=1/f'(a)$ Because continuity, when you are sending $h$ to zero, you send $k$ to zero too and that authorizes you to say that that limit is, so said, automatically happening :) – 2017-02-28
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    I think I can explain it with some algebra. I'll do if you need it. – 2017-02-28
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    @RafaBudría please do. I still haven't been able to figure out what happened. – 2017-02-28

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I'll try to answer satisfactorily. The point is that the second limit is not deduced from the first. Simply the derivative of $f$ at $a$ exist by hypothesis!! The manipulation is not over the limits, but over the variables into the limits, so is, $h$ and $k$.

You wrote this::

$$k = f^{-1}(b + h) - f^{-1}(b)$$

$k$ is a function of $h$, so is, $k=k(h)$. And because $f^{-1}$ is continuous:

$$\lim_{h\to 0}k(h) = \lim_{h\to 0}(f^{-1}(b + h) - f^{-1}(b))=0$$

$$\lim_{h \to 0} \dfrac{k(h)}{f(a + k(h)) - f(a)}=\lim_{k \to 0} \dfrac{k}{f(a + k) - f(a)}=$$

$$=\lim_{k \to 0} \frac{1}{\dfrac{f(a + k) - f(a)}{k}}=\frac{1}{f'(a)}$$

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    Thanks for the response. I understand your answer, but I'm not sure what you did with $=\lim_{k \to 0} \frac{1}{\dfrac{f(a + k) - f(a)}{k}}=\frac{1}{f'(a)}$. I think it is clearer to explain it as follows: $\lim_{h \to 0} \dfrac{k}{f(a + k) - f(a)} = \lim_{h \to 0} \dfrac{k}{b + h - b} = \lim_{h \to 0} \dfrac{k}{h} = \lim_{h \to 0} \dfrac{k}{f^{-1}(b + h) - f^{-1}(b)}$ where $f^{-1}(b + h) - f^{-1}(b) = k$. This explains how we got the denominator of $\lim_{k \to 0} \dfrac{f(a + k)) - f(a)}{k}$. – 2017-03-01