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Let $f : [1, \infty) \to [1, \infty)$ be a function such that $f(x) = x(1 + \ln x)$. Prove that $f$ is bijective and then calculate: $$\lim_{x \to \infty} \frac{f^{-1}(x) \ln x}{x}$$

I have no difficulties in proving that $f$ is bijective, but I can't calculate the limit. I've tried using l'Hospital's rule but got nothing meaningful.

Thank you in advance!

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    I'd like to know this too, I'm not sure, but maybe rewrite your limit in the inverse form, and then solve for $y \to \infty$? – 2017-02-06

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Since $f$ is bijective then there exists a unique $b$ such that $x=f(b)$ and if $x\to\infty$ then so does $b\to\infty$ and using that $f^{-1}(f(b))=b$ you're looking at $$\lim_{b\to\infty}\frac{b\ln f(b)}{f(b)}=\frac{b(\ln(b(1+\ln b)))}{b(1+\ln b)}=\frac{\ln(b)+\ln(1+\ln b)}{1+\ln b}{=^{L'h}}\lim_{b\to\infty}\frac{\frac{1}{b}+\frac{1}{b(1+\ln b)}}{\frac{1}{b}}=\lim_{b\to\infty}\frac{1+\frac{1}{1+\ln b}}{1}=1$$