$\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} $

Question

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.

Answer 1

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