This is exercise 5.2.12 from Understanding Analysis 2nd ed.
If $f : [a,b] \rightarrow \mathbb{R}$ is one-to-one , then there exists an inverse function $f^{-1}$ defined on the range of $f$ given by $f^{-1}(y) = x$ where $y=f(x)$. I previously went through the proof that if $f$ is continuous on $[a,b]$, then $f^{-1}$ is continuous on its domain.
Assume that $f$ is differentiable on $[a,b]$ with $f'(x) \neq 0$ for all $x \in [a,b]$. Show $f^{-1}$ is differentiable with $$(f^{-1})'(y) = \frac{1}{f'(x)}$$ where $y=f(x)$.
My work so far: Let $d = f(c)$ for an arbitrary $c \in [a,b]$. We have $$(f^{-1})'(d) = lim_{y \rightarrow d} \frac{f^{-1}(y) - f^{-1}(d)}{y-d}.$$
We can substitute in $f(x)$ for $y$ and $f(c)$ for $d$ to give $$(f^{-1})'(d) = (f^{-1})'(f(c))= lim_{f(x) \rightarrow f(c)} \frac{x - c}{f(x)-f(c)}.$$
It is at this point that I am stuck. All I need to do is show I can replace $f(x) \rightarrow f(c)$ with $x \rightarrow c$ in the limit. If I can do that, then (using the algebraic rules for limits) I would have $$lim_{x \rightarrow c} \frac{x - c}{f(x)-f(c)} = \frac{lim_{x \rightarrow c} 1}{lim_{x \rightarrow c} \frac{f(x) - f(c)}{x-c}} = \frac{1}{f'(c)}.$$
Just having trouble justifying the replacement of $f(x) \rightarrow f(c)$ with $x \rightarrow c$.