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Is the following limit computation correct:$$a = \lim\limits_{x\rightarrow 1} \exp\left\{\frac{W_{-1}\left(x\ln(x)\right)}{x}\right\} = \exp\left\{\frac{W_{-1}\left(1\cdot 0\right)}{1}\right\} = \exp(-\infty) = 0$$

More generally, when can we write:$$\lim\limits_{x\rightarrow x_0} W(x) = W\left(\lim\limits_{x\rightarrow x_0} x\right) = W(x_0)$$

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  • $\begingroup$ The limit is $-\infty$ since $x\log(x)\to 0$ when $x\to 1$. Look at the expansion in the Wikipedia page. $\endgroup$ – Claude Leibovici Dec 4 '18 at 10:48
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The principal branch of the Lambert function (considered as function of a real variable, which seems to be the case in the question) is continuous on $[-1/e,\infty)$, so the answer is yes if $x_0\in[-1/e,\infty)$ (limit from the right if $x_0=-1/e$.)

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  • $\begingroup$ Thank you for answering, I am using the second real branch in my computation where it is defined and seems to be continuous. So continuity is an enough criterion to compute the limit as this? $\endgroup$ – jlandercy Dec 4 '18 at 10:23
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    $\begingroup$ Yes, that is the definition of continuity. $\endgroup$ – Julián Aguirre Dec 4 '18 at 10:34
  • $\begingroup$ I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh. $\endgroup$ – jlandercy Dec 4 '18 at 10:40

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