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.


4 Answers 4


We have,

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


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


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

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.

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


  • 4
    $\begingroup$ How is that a "hint"? $\endgroup$
    – Clement C.
    Jan 30, 2017 at 3:15
  • $\begingroup$ Utilize the derivative of f(a) $\endgroup$ Jan 30, 2017 at 3:18
  • 2
    $\begingroup$ On your second application of L'Hôpital's, you appear to say that $\lim_x f(x)+\frac{g(x)}{h(x)}=\lim_x f(x)+\frac{g'(x)}{h'(x)}$ if $\frac{g(x)}{h(x)}$ is indeterminate, but doesn't that depend on $\lim_x f(x)$ and $\lim_x\frac{g(x)}{h(x)}$ existing independently? That's not clear here... $\endgroup$ Jan 30, 2017 at 4:52
  • 3
    $\begingroup$ @StevenStadnicki I use the addition rule for limits because $\lim_{a \to \infty} f'(a) \ln a$ and $\lim_{a \to \infty} \frac{f(a)}{a}$ both exist. The second existing should be pretty clear. As for the first substituting our expression for the derivative we have $\lim_{a \to \infty} \frac{\ln a}{2 +\ln f(a)}$ and continuing on just as I did in the answer we get $1$. So both limits exist, and thus I can use the addition rule. $\endgroup$ Jan 30, 2017 at 5:23
  • 2
    $\begingroup$ Believe it or not but this is a truly smart, efficient and rigorous use of L'Hospital's Rule and +1 for the same. $\endgroup$
    – Paramanand Singh
    Jan 30, 2017 at 9:27

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

  • $\begingroup$ Looks like there is no limit to getting simpler and simpler answers. +1 for the shortest and simplest answer so far. $\endgroup$
    – Paramanand Singh
    Jan 30, 2017 at 15:38
  • $\begingroup$ Yes, this is the way I though the exercise... seems more clear than the other both answers (well, it seems more elementary). $\endgroup$
    – Masacroso
    Jan 30, 2017 at 16:20

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 $$\lim_{a \to \infty} \frac{f(a)\ln a}{a}\sim \lim_{a \to \infty}\frac{\ln(a)}{\ln(ae)}=\lim_{a \to \infty}\frac{\ln(a)}{\ln(a)+1}=1$$


You should try to look for an equivalent of $f(a)$ or $\ln(f(a))$:

First as you said f is an increasing function, and: $\lim f = + \infty$

So you have: $f(a)[1+\ln(f(a))]=a \implies f(a)\ln(f(a))$~$a$

Here it means: $f(a)\ln(f(a))= a + o(a) = a[1+ o(1)]$

You might want to consider the $\ln$ on both sides


I'll detail o() and ~ notations so that you can understand here why it can be nice to use it:

Let f and g be two real valued functions:

  • f(x) = o(g(x)) when $x \rightarrow + \infty $ means: $\forall \epsilon >0 , \exists A \in R :x>A \implies |f(x)|< \epsilon|g(x)|$

    If g never cancels, it's equivalent to: $\lim \frac{f(x)}{g(x)} = 0 , x \rightarrow + \infty$

Likewise you can define this notion when $x \rightarrow a , a \in R$ if for instance g diverges in a. f is said to be negligible compared to g.

  • f(x) ~ g(x) when $x \rightarrow + \infty$ means : $f(x) = g(x) + o(g(x))$

When g never cancels it's the same as: $\lim \frac{f(x)}{g(x)} = 1 , x \rightarrow + \infty$

f is then said to be equivalent to g in $+ \infty$.

Likewise, the notion extends to the case where $x \rightarrow a , a \in R$.

Now since you're not familiar with this i can show you why it's nice to use it sometimes, like here:

$f(a)[1+\ln(f(a))]=a , f(a) \rightarrow +\infty$ when $a \rightarrow +\infty$:

1 is then negligible compared to ln(f(a)), and:

$1= \frac{f(a)[1+\ln(f(a))]}{a} = \lim_{a\rightarrow +\infty} \frac{f(a)\ln(f(a))}{a}$

So you have: $f(a)\ln(f(a)) \sim_{a\rightarrow +\infty} a$

So using the above definition you end up with what i had :

$f(a)\ln(f(a))= a + o(a) = a[1+ o(1)] $

where $o(1)$ is a function that verifies: $o(1) \rightarrow 0$ when $a\rightarrow +\infty$

See the nice thing is that you can manipulate an equation easily now, so if you use $\ln$ on both sides:

$\ln[f(a)\ln(f(a))] = \ln(f(a))+ \ln(\ln(f(a))) = \ln(a) + \ln[1+o(1)]$

You know that $\ln$ is continuous, and $\ln(1)=0$ so $\ln[1+o(1)]\rightarrow_{a\rightarrow +\infty} \ln(1)=0 \implies \ln[1+o(1)]=o(1)$ , when $a\rightarrow +\infty$

So you get: $\ln(f(a))+ \ln(\ln(f(a))) = \ln(a) + \ln[1+o(1)] = \ln(a) + o(1)$

Finally, since: $\frac{\ln x}{x}\rightarrow 0$ , when $x\rightarrow +\infty$ :

$\frac{\ln(\ln(f(a)))}{\ln f(a)} \rightarrow 0$ when $a\rightarrow +\infty$

So the equality writes:

$\ln(f(a))+ \ln(\ln(f(a))) =\ln(f(a))+o(\ln(f(a))) = \ln(a)+o(1) \implies \ln(a) = \ln(f(a)) + o(\ln(f(a))) -o(1) =\ln(f(a)) +o(\ln(f(a))) $

Since o(1) is also negligible compared to $\ln(f(a))$ so it is a $o(\ln(f(a)))$

Hence you have: $\ln(a)=\ln(f(a)) +o(\ln(f(a)))$ and this means exactly here that:

$$\lim_{a\rightarrow +\infty} \frac{\ln(a)}{\ln(f(a))}=1$$

Now you can find the limit... It's a bit long sorry but hopefully you will see why this can be powerful once you're comfortable with the notions :)

  • $\begingroup$ Are you familiar with the o() and ~ notations? $\endgroup$
    – mvggz
    Jan 30, 2017 at 3:28
  • $\begingroup$ I am afraid I am not aware of the notations.Can you please explain them briefly or suggest a link where I may get to know about them. $\endgroup$
    – Navin
    Jan 30, 2017 at 3:36
  • $\begingroup$ Hello, you can check this wikipedia page: en.wikipedia.org/wiki/Big_O_notation I'll edit the answer to explain this to you , it's really simple but practical $\endgroup$
    – mvggz
    Jan 30, 2017 at 13:51
  • 2
    $\begingroup$ Your answer is correct/rigorous apart from a minor glitch (which I will consider as a typo). This shows that one can establish seemingly complicated limits (which perhaps require multiple applications of L'Hospital's Rule in very smart and non-obvious manner, like in answer from Ahmed Attaalla) via the use of standard limits like $(\log x) /x\to 0$ combined with certain algebraic manipulation. +1 for the excellent answer. I have fixed the typo BTW. $\endgroup$
    – Paramanand Singh
    Jan 30, 2017 at 15:32
  • $\begingroup$ Cheers man :) . Yes I like to use this kind of approach when I have the chance, it's actually less painful for the brain and concise (if you're familiar with the notions). I hope OP will have a use for it $\endgroup$
    – mvggz
    Jan 30, 2017 at 15:42

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