Limit containing an unknown inverse function

Question

If$$ f(x) = x + \sin(x),\ g(y) =f^{-1}(y),$$ find$$ \lim_{y\to 0} \frac{2g(y)-y}{y^2}. $$ Can you help me with this please? Thank you.

Answer 1

Notice that $f$ and its inverse $g$ are a strictly increasing differentiable function such that $g'(y)=1/f'(x)$ with $y=f(x)$ (take a look at inverse functions and differentiation).

Now $g(0)=f(0)=0$ and by using L'Hopital we have that $$\lim_{y\to 0} \frac{2g(y)-y}{y^2}=\lim_{y\to 0} \frac{2g'(y)-1}{2y} =\lim_{x\to 0} \frac{2/f'(x)-1}{2f(x)}.$$ Can you take it from here?