[SOLVED] Limit when Hopital's rule fails

I have a function $\displaystyle f(x) = g^{-1}(h(x))$ and I'm looking for $\displaystyle \lim_{x \rightarrow x_0^+} f'(x)$

I know the following about the functions g and h:

- $\displaystyle h(x_0)=0$ and $\displaystyle g(0) = 0$ so $\displaystyle f(x_0)=0$
- $\displaystyle g \geq 0, h \geq 0$
- $\displaystyle g'(0)=0, h'(x_0)=0$
- $\displaystyle g' > 0$ for $\displaystyle x \neq x_0$, $\displaystyle h' > 0$ for $\displaystyle x<x_0$ and $\displaystyle h'<0$ for $\displaystyle x>x_0$
- $\displaystyle g'' > 0, h'' > 0$ and finite at $\displaystyle x_0$
- $\displaystyle h''' = 0, g'''$ finite at $\displaystyle x_0$

We have $\displaystyle f'(x)= \frac{h'(x)}{g'(f(x))}$. At $\displaystyle x_0$, both numerator and denominator are 0. Applying l'Hopital's rule, we obtain

$\displaystyle \lim_{x \rightarrow x_0^+} f'(x) = \lim_{x \rightarrow x_0^+} \frac{h''(x)}{f'(x) g''(f(x))} = \lim_{x \rightarrow x_0^+} \frac{h''(x) g'(f(x))}{h'(x) g''(f(x))} $.

Applying l'Hopital's rule again, we're back to where we started:

$\displaystyle \lim_{x \rightarrow x_0^+} \frac{h'''(x) g'(f(x))+h''(x) f'(x) g''(f(x))}{h''(x) g''(f(x)) + h'(x) f'(x) g'''(f(x))} = \lim_{x \rightarrow x_0^+} f'(x). $

There's probably a simpler way of presenting this problem, sorry about that... If anyone knows how this kind of circularity problem is called and/or how to deal with it, I will be very grateful!