Find $\lim_{x\to \infty} \ln(\exp(\operatorname{LmW}(x))+1)(\exp(\operatorname{LmW}(x))+1) - x - \ln(x)$

Find $\lim_{x\to \infty} \ln(e^{\operatorname{LambertW}(x)}+1)(e^{\operatorname{LambertW}(x)}+1) - x - \ln(x)$

Where the $LambertW$ function is defined here : http://en.wikipedia.org/wiki/Lambert_W

How to do this ?

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I see no difference after the edit ? Only the + for infinity has been removed , which ironicly I would consider an edit if it was left out. Maybe I wrote find without capital. Im not complaining Im just curious what it was. – mick Oct 4 '12 at 14:36
As for the suggestion of tag special function , that special function can be easily substituted away. So Im unsure about that. – mick Oct 4 '12 at 14:37
The special function tag might attract other people to the question. Don't be mad, edits are made to help the OP as well as the readers. For example $$\lim_{x\rightarrow \infty} \big\{(e^{W(x)} + 1)\ln\big(e^{W(x)}+1\big) - x - \ln(x)\big\}$$ is far more readable. – Pragabhava Oct 4 '12 at 14:42
@Pragabhava : Hmm the $e$ instead of $exp$ is worth consideration. But not everybody knows $W(x)$ although I could define it. But Im more concerned about leaving out the +. – mick Oct 4 '12 at 15:01
@Pragabhava : Besides Im not mad. Extravert Psychology : Axiom 1 : 'not complaining' <-> not mad. Ok I made that up :) – mick Oct 4 '12 at 15:04

Maple says $$\lim_{x \to \infty}\left[\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)} \operatorname{ln} \biggl(\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)}\biggr) - x - \operatorname{ln} (x)\right] = \infty$$ if that's what you mean. In fact, $$\lim_{x \to \infty}\frac{1}{x}\left[\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)} \operatorname{ln} \biggl(\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)}\biggr) - x - \operatorname{ln} (x)\right] = e-1$$
OK, on this limit $$\lim_{x\to\infty}\left[\Bigl(\operatorname{e} ^{W (x)} + 1\Bigr) \operatorname{ln} \Bigl(\operatorname{e} ^{W (x)} + 1\Bigr) - x - \operatorname{ln} (x)\right]$$ Maple says "Too many levels of recursion". So I used $e^{W(x)}=x/W(x)$ and then let $x=e^y$. (Also use: $\ln W(e^y) = y - W(e^y)$.) Now Maple says $$\lim_{y\to\infty} \;\left[\frac{\operatorname{e} ^{y} \operatorname{ln} \bigl(\operatorname{e} ^{y} + W \bigl(\operatorname{e} ^{y}\bigr)\bigr)}{W \bigl(\operatorname{e} ^{y}\bigr)} - \frac{\operatorname{e} ^{y} y}{W \bigl(\operatorname{e} ^{y}\bigr)} + \operatorname{ln} \Bigl(\operatorname{e} ^{y} + W \bigl(\operatorname{e} ^{y}\bigr)\Bigr) - 2 y + W \bigl(\operatorname{e} ^{y}\bigr)\right] = -\infty$$ which is claimed as your answer...