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?
 A: 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 . 
A: Since $f$ is monotone increasing for $x$ large (for all $x$ even), we can substitute $f(y)$ for $x$ in the limit as follows.
$$
\lim_{x \to \infty} \frac{f^{-1}(x)}{\ln(x)} = \lim_{y \to \infty} \frac{y}{\ln(f(y))}
$$
Now,
\begin{align}
\lim_{y \to \infty} \frac{y}{ln(f(y))} &= \lim_{y \to \infty} \frac{y}{\ln(e^y + y^3 - y^2 + y)} \\
&=\lim_{y \to \infty} \frac{y}{y + \ln(1 + e^{-y}y^3 - e^{-y}y^2 + e^{-y}y)}. \\
\end{align}
As $e^{-y}$ times any polynomial goes to $0$ as $y \to \infty$ we see that the limit is $1$.
A: Expanding my comment into an answer.

The derivative $$f'(x) =e^x+\text{ a polynomial in } x$$ clearly tends to $\infty$ as $x\to\infty $ and hence is positive in some neighborhood of type $(a, \infty) $ and therefore $f$ is strictly increasing in that neighborhood. This implies that $f$ is invertible in the same neighborhood.
Further $f(x) \to\infty $ as $x\to\infty $ hence $t=f^{-1}(x)$ also tends to $\infty$ as $x\to\infty $. Putting $t=f^{-1}(x)$  the limit in question reduces to $$\lim_{t\to\infty} \frac{t} {\log f(t)} $$ and clearly $$\log f(t) = \log (e^t+p(t)) =t+\log(1+p(t)e^{-t})$$ where $p(t) $ is some polynomial and therefore $p(t) e^{-t} \to 0$. It follows that $$\frac {t} {\log f(t)} =\dfrac{1}{1+\dfrac{\log(1+p(t)e^{-t})}{t}}\to\frac{1}{1+0}=1$$ as $t\to\infty $.

The question is specifically intended to check whether one is trying to figure out the inverse of $f$ in explicit manner or not. The limit evaluation does not really require the inverse of $f$ and the idea is to transform the limit in question to one dealing with the function $f$. 
A: For $x \to \infty, f(x) \sim e^x$ so $f^{-1}(x) \sim \ln(x)$. So the answer is $1$.
