TITLE: Finding out the limit $\lim_{a \to \infty} \frac{f(a)\ln a}{a}$
QUESTION [12 upvotes]: For any real number $a \geq 1$ let $f(a)$ denote the real solution of the equation $x(1+\ln x)=a$ then the    question is to find out $$ \lim_{a \to \infty} \frac{f(a)\ln a}{a}$$.
It is clear that if we denote $h(a)$ by $h(a)=a(1+\ln a)$ then $f(a)$ is the inverse function of $h(a)$. Also $f(a)$ is increasing function in its domain. Also the limit persuades using lhospital's but I cannot see how to apply it here. Thanks.

REPLY [4 votes]: The function $g(x)=x(1+\ln x)$ defined over $[1,\infty)$ has derivative
$$
g'(x)=1+\ln x+1=2+\ln x>0
$$
and $\lim_{x\to\infty}g(x)=\infty$,
so the function is increasing and therefore it has an inverse function defined over $[g(1),\infty)=[1,\infty)$. Its inverse is exactly the function $f$ you have to analyze the behavior of.
Now you can use the substitution $a=g(x)$ so the limit becomes
$$
\lim_{x\to\infty}\frac{x\ln(g(x))}{g(x)}=
\lim_{x\to\infty}\frac{x\ln\bigl(x(1+\ln x)\bigr)}{x(1+\ln x)}=
\lim_{x\to\infty}\frac{\ln x}{1+\ln x}+
\lim_{x\to\infty}\frac{\ln(1+\ln x)}{1+\ln x}=1
$$