$\lim_{x \to 0} \frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)} = 1$ for any $f,g \in C^1$ that are tangent to $\text{id}$ at $0$ with some simple condition Theorem
For any real functions $f,g \in C^1$ such that $f(0) = g(0) = 0$ and $f'(0) = g'(0) = 1$ and $x$ is strictly between $f(x)$ and $g(x)$ for any $x \ne 0$:
  $f,g$ are invertible on some open neighbourhood of $0$
  $\dfrac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)} \to 1$ as $x \to 0$
Questions
What is the simplest proof you can think of? I've given mine below.
Motivation
This theorem was inspired by user8286's solution to a special case, and I wanted to find weaker conditions on the functions under which the limit in his proof would hold.
 A: [Edit: The first half was simplified thanks to Paramanand Singh.]
Proof
For any real functions $f,g \in C^1$ such that $f(0) = g(0) = 0$ and $f'(0) = g'(0) = 1$ and $x$ is strictly between $f(x)$ and $g(x)$ for any $x \ne 0$:
  Let $d > 0$ such that $f'(x) > 0$ for any $x \in (-d,d)$
  Then $f$ is strictly increasing on $(-d,d)$ by Mean value theorem
  Similarly $g$ is strictly increasing on some open neighbourhood of $0$
  Therefore $f,g$ are invertible on some open neighbourhood of $0$ because $f,g \in C^1$
  WLOG $f(x) > x > g(x)$ for any $x \ne 0$ because $f,g$ are continuous and by symmetry
  As $x \to 0$:
    $\dfrac{f(x)-x}{x-f^{-1}(x)} = f'(a)$ for some $a \in (f^{-1}(x),x)$ by Mean value theorem
    $\dfrac{x-g(x)}{g^{-1}(x)-x} = g'(b)$ for some $b \in (x,g^{-1}(x))$ by Mean value theorem
    $f'(a) \to f'(0) = 1$ because $x \to 0$ and $f^{-1}(x) \to 0$
    $g'(b) \to g'(0) = 1$ because $x \to 0$ and $g^{-1}(x) \to 0$
    Therefore $\dfrac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)} \to 1$ because $\dfrac{p+q}{r+s}$ is between $\dfrac{p}{r}$ and $\dfrac{q}{s}$ for any $p,q,r,s > 0$
