For $f(x) = e^x + x^3 - x^2 + x$ find the limit $\lim\limits_{x\to \infty} \frac{f^{-1}(x)}{\ln x}$. I have the function:

$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm} f(x) = e^x + x^3 -x^2 + x$$

and I have to find the limit:

$$\lim\limits_{x \to \infty} \frac{f^{-1}(x)}{\ln x}$$

(In the first part of the problem, I had to show that the function is strictly increasing and invertible. I don't know if that's relevant to this, since I could show that the function is invertible, but I can't find the inverse.)

So this is what I tried:

I showed

$$\lim_{x \to \infty} f(x) = \infty$$

and so I concluded that $\lim\limits_{x \to \infty} f^{-1}(x) = \infty$. I'm not sure if this is correct, it might be wrong. But if would be right, then we could use l'Hospital, knowing that:

$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$$

but after trying to use all of this on paper, I got nowhere. It just complicated thing a lot more.

So how should I solve this limit?
Hint : $$\forall x\geq 1\quad 2e^x \geq e^x+x^3-x^2+x\geq e^x+1$$

So $$\ln(\frac{x}{2})\leq f^{-1}(x)\leq \ln(x-1)$$

So we deduce that your limit is a constant by the sandwich theorem .
For $x \to \infty, f(x) \sim e^x$ so $f^{-1}(x) \sim \ln(x)$. So the answer is $1$.