$ \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.
 A: The fact that $\lim_{x \to  0^+} \frac{f(2x^2)}{\sqrt {3}x^2} = 1$ implies that the $\lim_{x \to 0^+}f(2x^2) = 0$.  Since $f$ is continuous, this in turn implies that $f(0) = 0$.
Since they give us the invertible function, presumably, $f$ must be invertible, which is to say that is at least one-to-one.  So, in particular, we have $f^{-1}(0) = 0$.  They don't explicitly tell us how "nice" $f^{-1}$ is, so I'm going to try getting by assuming only that $f^{-1}$ is continuous.

Now: to deal with the limit in question, we will make the substitution 
$x = f(y)$
to give us the new limit
$$
\lim_{x \to  0^+} \frac{f(f(x)) }{f^{-1}(x)} = 
\lim_{y \to 0^+} \frac{f(f(f(y))) }{y}
$$
Try to justify to yourself that these two limits are the same.  We can now rewrite the second limit as
$$
\lim_{h \to 0^+} \frac{f(f(f(0+h))) - f(f(f(0)))}{h} = 
\left. \frac {d}{dx^+}[f(f(f(x)))] \right|_{x = 0}
$$
Now it's clear why we wanted to have $f \in \mathcal C^1$.  Remember, the chain rule gives us
$$
\frac {d}{dx}[f(f(f(x)))] = f'(f(f(x)))\cdot f'(f(x))\cdot f'(x)
$$
All that's left, then, is to find $f'(0)$.

Note that applying L'Hôpital's rule to the limit given to us yields
$$
1 = \lim_{x \to  0^+} \frac{f(2x^2)}{\sqrt {3}x^2}
= \lim_{x \to 0^+} \frac{f'(2x^2) \cdot 4x}{2\sqrt {3}x}
= \lim_{x \to 0^+} \frac{2}{\sqrt{3}} f'(2x)
= \frac 2{\sqrt 3} f'(0)
$$
Or, alternatively, taking the substitution $u = 2x^2$ gives us
$$
1 = \lim_{x \to 0^+} \frac{f(u)}{(\sqrt{3}/2)u} = \frac 2{\sqrt 3} \lim_{u \to 0^+} \frac{f(u)}u = \frac 2{\sqrt 3} f'(0)
$$
A: $$\displaystyle\lim_{x \to  0^+} \frac{f(2x^2)}{\sqrt {3}x^2} = 1 $$ implies $$f(x) = \frac{\sqrt 3}{2} x + \cdots, \quad f^{-1}(y) = \frac2{\sqrt 3}y+\cdots .$$  then we also have $$f(f(x)) = \frac34 x+\cdots $$
therefore $$\displaystyle \lim_{x \to  0^+} \frac{f(f(x)) }{f^{-1}(x)}=\lim_{x \to 0+}\frac{3x/4+\cdots}{2x/\sqrt 3+\cdots} =\frac{3\sqrt 3}8.$$
A: 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. $(f^{-1})'(0)=\frac1{f'(0)}$
