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?

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 $\frac{y}{\ln x}\to 1$.
