Lagrange/Cauchy mean value theorem Let $\,\,f,g\in C^2([0,1])$ such that $f'(0)g''(0)-f''(0)g'(0)\neq 0$ and $g'(x)\neq 0$ for all $x\in (0,1)$. Let $\theta(x)$ be a real number such that
$$\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(\theta(x))}{g'(\theta (x))}.$$
What can I say about $\lim_{x\rightarrow 0^+} \frac{\theta(x)}{x}$?
 A: We have:
$$ f(x) = f(0)+f'(0) x +\int_{0}^{x}f''(t)(x-t)\,dt $$
$$ f'(x) = f'(0)+\int_{0}^{x}f''(t)\,dt $$
hence:
$$\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(0)+\int_{0}^{x}f''(t)(1-t/x)\,dt}{g'(0)+\int_{0}^{x}g''(t)(1-t/x)\,dt}=\frac{f'(0)+\int_{0}^{\theta(x)}f''(t)\,dt}{g'(0)+\int_{0}^{\theta(x)}g''(t)\,dt}.\tag{1}$$
Assuming that over the interval $I_x=[0,x]$ both $f''$ and $g''$ have small variation ($\leq\varepsilon$), since $\theta(x)\in I_x$ we have that $\int_{0}^{\theta(x)}f''(t)$ behaves like $\theta(x)\,f''(0)$ while $\int_{0}^{x}f''(t)(1-t/x)\,dt$ behaves like $\frac{x}{2}f''(0)$, hence 
$$\frac{f'(0)+\frac{x}{2}f''(0)}{g'(0)+\frac{x}{2}g''(0)} =\frac{f'(0)+\int_{0}^{x}f''(t)(1-t/x)\,dt}{g'(0)+\frac{x}{2}g''(0)}=$$$$ =\frac{f'(0)+\int_{0}^{\theta(x)}f''(t)\,dt}{g'(0)+\frac{x}{2}g''(0)} = \frac{f'(0)+\theta(x)\,f''(0)}{g'(0)+\frac{x}{2}g''(0)}.\tag{2}$$
Collecting...
$$\frac{f'(0)+\frac{x}{2}f''(0)}{g'(0)+\frac{x}{2}g''(0)} = \frac{f'(0)+\theta(x)\,f''(0)}{g'(0)+\frac{x}{2}g''(0)} \Rightarrow \frac{x}{2}f''(0) = \theta(x)\,f''(0) .\tag{3}$$
And taking limit we get that
$$\lim_{x\to 0^+}\frac{\theta(x)}{x}=\frac{1}{2}.$$
