Finding out the limit $\lim_{a \to \infty} \frac{f(a)\ln a}{a}$

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.

We have,

$$f(a)(1+\ln f(a))=a$$

Hence,

$$f'(a)(1+\ln f(a))+f(a)\frac{1}{f(a)}f'(a)=1$$

So,

$$f'(a)=\frac{1}{2+\ln f(a)}$$

By l'Hopitals what we are interested in is,

$$\lim_{a \to \infty} \left( f'(a) \ln a+\frac{f(a)}{a} \right)$$

Again by l'Hopitals on the second limit because it is clear $f(a) \to \infty$ as $a \to \infty$, the use of the addition rule for limits will be justified by the end of this answer.

$$\lim_{a \to \infty} f'(a) \ln a+\lim_{a \to \infty} f'(a)$$

$$=\lim_{a \to \infty} f'(a)\ln a$$

Now substitute,

$$f'(a)=\frac{1}{2+\ln f(a)}$$

To get,

$$=\lim_{a \to \infty} \frac{\ln a}{2+\ln f(a)}$$

Utilize l'Hopitals

$$=\lim_{a \to \infty} \frac{f(a)}{af'(a)}$$

Substitute our expression for the derivative back in.

$$=\lim_{a \to \infty} \frac{f(a)(2+ \ln f(a))}{a}$$

Utilize l'Hopitals

$$=\lim_{a \to \infty} \left(f'(a)(2+\ln f(a))+f'(a) \right)$$

Substitute our expression for the derivative back in.

$$=\lim_{a \to \infty} (1+f'(a))$$

$$=1$$

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 $$ \lim_{x\to\infty}\frac{x\ln(g(x))}{g(x)}= \lim_{x\to\infty}\frac{x\ln\bigl(x(1+\ln x)\bigr)}{x(1+\ln x)}= \lim_{x\to\infty}\frac{\ln x}{1+\ln x}+ \lim_{x\to\infty}\frac{\ln(1+\ln x)}{1+\ln x}=1 $$

For the time being, this is totally off-topic.

$$x(1+\ln x)=a\implies (xe)\ln(xe)=ae \implies x=\frac{a}{W(e a)}$$ where appears Lambert function. This makes $$\frac{f(a)\ln (a)}{a}=\frac{\ln (a)}{W(e a)}$$ In the Wikipedia page, you will notice that, when $t\to \infty$, $W(t)\sim \ln(t)$ which makes $$\lim_{a \to \infty} \frac{f(a)\ln a}{a}\sim \lim_{a \to \infty}\frac{\ln(a)}{\ln(ae)}=\lim_{a \to \infty}\frac{\ln(a)}{\ln(a)+1}=1$$