$\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(\nu(x))}{g'(\nu(x))} $ ,the value of the limit: $\lim_{x \to 0^+} \frac{\nu(x)}{x} $ Good evening,
I thought a lot about this issue.
I think I have to apply Lagrange, Taylor.
Can someone  help me to calculate this limit?
$$f,g \in C^2 [0,1]:   \\ f'(0)g''(0) \ne f''(0) g'(0) \\ g'(x) \ne 0, \forall x \in (0,1)  \\ \nu(x) \text{ is a real number }: \\ \frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(\nu(x))}{g'(\nu(x))} \\ \lim_{x -> 0^+} \frac {\nu(x)}{x} $$
My reasoning, using Taylor:
$ f(x)=f(0)+xf'(0)+x^2 \frac{f''(0)}{2} \\
 g(x)=g(0)+xg'(0)+x^2 \frac{g''(0)}{2} \\
\frac{f'(0)+\frac{f''(0)}{2}x}{g'(0)+\frac{g''(0)}{2}x}=\frac{f'(\nu(x))}{g'(\nu(x)) } \\
\frac{f'(0)+\frac{f''(0)}{2}x}{g'(0)+\frac{g''(0)}{2}x}=\frac{f'(\nu(0))+\nu'(0)f''(\nu(0))x}{g'(\nu(0))+\nu'(0)g''(\nu(0))x }
$
Can you give me a hint to continue the reasoning? Is there any mistake?
Thanks. 

$\nu(0)=0$ because $0\leq \nu(x)\leq x$.  Then cross-multiply, and take $O(x)$ terms.
From your last line, everything is evaluated at zero:
$$(f'+f''x/2)(g'+g''\nu'x)\approx(f'+f''\nu'x)(g'+g''x/2)\\
f'g'+(f''g'+2f'g''\nu')x/2+Ax^2\approx f'g'+(2g'f''\nu'+f'g'')x/2+Bx^2\\
f''g'+2f'g''\nu'=2g'f''\nu'+f'g''\\
\nu'=1/2$$
