Limit of $ \frac{f(x) f''(x)}{(f'(x))^{2}}$ at $\alpha$ with $f(\alpha)= f'(\alpha)= f''(\alpha)=0$ and $f'''(\alpha) \not= 0$ via L'hopital

Question

Suppose $f$ is a $C^{4}$ function with $f(\alpha)= f'(\alpha)= f''(\alpha)=0$ and $f'''(\alpha) \not= 0$. I wish to compute the following limit:

$$ \lim_{x \to \alpha} \frac{f(x) f''(x)}{(f'(x))^{2}}.$$

The result I get by repeatedly applying L'hopital is $\frac{1}{2}$, but if I assume $f \in C^{5}$ I am able to compute $\frac{2}{3}$.

Which one is the right value?

Answer 1

You can use Taylor expansion here. $$ f(x)=f'''(\alpha)(x-\alpha)^3/6+o(x-\alpha)^3, $$ $$ f'(x)=f'''(\alpha)(x-\alpha)^2/2+o(x-\alpha)^2 $$ and $$ f''(x)=f'''(\alpha)(x-\alpha)+o(x-\alpha) $$ Just replace everything by its expansion and you get $2/3$.