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