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

Question

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 ?

Answer 1

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, $$ \lim_{x \to \infty}\frac{1}{x}\left[\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)} \operatorname{ln} \biggl(\operatorname{e} ^{\bigl(LambertW (x) + 1\bigr)}\biggr) - x - \operatorname{ln} (x)\right] = e-1 $$

edit Oct 4

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...