Evaluating $\lim_{x\to\infty}\frac{f^{-1}(x)}{\ln(x)}$, where $f(x)=e^x+x^3-x^2+x$ 
$$f(x)=e^x+x^3-x^2+x$$
What is the following?
$$\lim_{x\to\infty}\frac{f^{-1}(x)}{\ln(x)}$$

Do I need to calculate the inverse of $f(x)$ or is there some other way that this limit can be solved?
 A: Since we want the limit at $\infty$, we may as well assume that the domain of $f$ is $(0,\infty)$. Note that $f'(x)=e^x+3x^2-2x+1=e^x+(\sqrt{3}x-\frac{1}{\sqrt{3}})^2+\frac{2}{3}$, so it is always positive on the domain. Hence $f$ is monotone and, viewed as $f:(0,\infty)\mapsto(2,\infty)$, is bijective.
Next observe that for any fixed $\delta>0$, $e^x<f(x)<e^{(1+\delta)x}$ for $x$ large enough (the cutoff point depending on $\delta$ of course). So, using monotonicity, for any $y\in (2,\infty)$ we have $e^x<y=f(x)<e^{(1+\delta)x}$ implying
\begin{equation*}
x<\log{y}<(1+\delta)x
\end{equation*}
On the other hand, observe that $f^{-1}$ is also strictly monotone (e.g. differentiate the identity $f(f^{-1}(x))=x$), so we have that $f^{-1}(e^x)<f^{-1}(y)<f^{-1}(e^{(1+\delta)x})$.
Claim: $x(1-\delta)<f^{-1}(e^x)$. Applying $f$ this is equivalent to $e^x>f(x(1-\delta))$, which can be checked the same way as before, for a possibly larger $x$.
Claim: $f^{-1}(e^{(1+\delta)x})<x(1+2\delta)$. Same as the previous claim.
Putting everything together, we have that
\begin{equation*}
x(1-\delta)<f^{-1}(y)<x(1+2\delta).
\end{equation*}
Therefore, for arbitrary $\delta>0$ we have that for any $y\in(2,\infty)$ large enough
\begin{equation*}
\frac{1-\delta}{1+\delta}<\frac{f^{-1}(y)}{\log{y}}<\frac{1+2\delta}{1},
\end{equation*}
so the limit is $1$.
A: Restrict $f(x)=e^x+p(x)$ to large $x$ so it is strictly increasing. Then by composing with $f^{-1}$ we have $x=e^{f^{-1}(x)}+p(f^{-1}(x))$, and by differentiation and rearrangement we get:
$${f^{-1}}'(x)=\frac{1}{e^{f^{-1}(x)}+p'(f^{-1}(x))}$$
Hence by L'hopital:
$$\lim_{x\to\infty}\frac{f^{-1}(x)}{\ln(x)}=
\lim_{x\to\infty}\frac{x}{e^{f^{-1}(x)}+p'(f^{-1}(x))}$$
Composed with $f(x)$
$$\lim_{x\to\infty}\frac{e^x+p(x)}{e^{x}+p'(x)}$$
Which will be $1$ because $p$ is a polynomial.
A: We have
$$ f'(x) = e^x + 3x^2 - 2x + 1 = e^x + 3(x-\frac{1}{3})^2 + \frac{2}{3} $$
so $f'(x) > 0$ for all $x$, and $f$ is a strictly increasing function. Then $f^{-1}$ is also strictly increasing, and since $\lim_{x \to +\infty} f(x) = +\infty$, we also know $\lim_{y \to +\infty} f^{-1}(y) = +\infty$.
If $f^{-1}(x) = t$, then $x = f(t)$. That is,
$$ x = e^t + t^3 - t^2 + t $$
So
$$ \frac{f^{-1}(x)}{\ln x} = \frac{t}{\ln f(t)} = \frac{t}{\ln(e^t + t^3 - t^2 + t)} $$
As $x \to +\infty$ and $t \to +\infty$, this is an $\infty/\infty$ form, so we can try L'Hopital's rule (multiple times):
$$ \begin{align*}
\lim_{x \to +\infty} \frac{f^{-1}(x)}{\ln x} &= \lim_{t \to +\infty} \frac{1}{\frac{e^t + 3t^2 - 2t + 1}{e^t + t^3 - t^2 + t}} 
= \lim_{t \to +\infty} \frac{e^t + t^3 - t^2 + t}{e^t + 3t^2 - 2t + 1} \\
&= \lim_{t \to +\infty} \frac{e^t + 3t^2 - 2t + 1}{e^t + 6t - 2} \\
&= \lim_{t \to +\infty} \frac{e^t + 6t - 2}{e^t + 6} \\
&= \lim_{t \to +\infty} \frac{e^t + 6}{e^t} = \lim_{t \to +\infty} 1 + 6e^{-t} = 1
\end{align*} $$
So the limit does in fact exist, with value $1$.
A: From $f'(x) = e^x + 3 x^2 - 2 x+1$ we deduce that $f'(x)\ge 3x^2 -2 x + 1>0$, hence $f$ is a bijection from ${\mathbb R}$ onto $f({\mathbb R}) = {\mathbb R}$ and we have $f^{-1}(x)\to +\infty$ when $x\to +\infty$. Let $y = f^{-1}(x)$, we have
\begin{equation}
\frac{\ln x}{y} = \frac{\ln(f(y))}{y} = \frac{y + \ln(1 + e^{-y}(y^3 -y^2+y))}{y} = 1 + \frac{1}{y}\ln(1 + e^{-y}(y^3 -y^2+y))
\end{equation}
Hence $\frac{\ln x}{y}\to 1$, hence $\frac{y}{\ln x}\to 1$.
