$ \lim_{x \to 0^+} \frac{f(f(x)) }{f^{-1}(x)}$ 
Suppose that $f \in \mathcal{C} ^1 \ ([0,1])$ and that $\displaystyle\lim_{x \to  0^+} \frac{f(2x^2)}{\sqrt {3}x^2} = 1$.
Find $\displaystyle \lim_{x \to  0^+} \frac{f(f(x)) }{f^{-1}(x)}$.

I don't know were to begin. They don't say if $\lim_{x \to  0^+} f(2x^2) = 0$, so is $f(2x^2) \sim \sqrt {3} x^2 $?
I also don't know how to deal with the inverse function.
Any help would be greatly appreciated, especially explaining the reasoning behind the method.

It is easy to verify that $f'(0)=\frac{\sqrt3}2$ and $f(0)=0=f^{-1}(0)=f(f(0))$
So , $\lim_{x\to0} \frac{f(f(x))}{f^{-1}(x)}$=$\lim_{x\to0} \frac{f(f(x))-f(f(0))}{x-0}\frac 1 {\frac{f^{-1}(x)-f^{-1}(0)}{x-0}}$=$\frac{(f'(0))^2}{(f^{-1})'(0)}$
and $(f^{-1})'(y)=\frac1{f'(x)}$ where $y=f(x)$ i.e. $(f^{-1})'(0)=\frac1{f'(0)}$
