# Verifying a limit with Lambert W function

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:$$\lim\limits_{x\rightarrow x_0} W(x) = W\left(\lim\limits_{x\rightarrow x_0} x\right) = W(x_0)$$

• The limit is $-\infty$ since $x\log(x)\to 0$ when $x\to 1$. Look at the expansion in the Wikipedia page. – Claude Leibovici Dec 4 '18 at 10:48

## 1 Answer

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 $$x_0=-1/e$$.)

• 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? – jlandercy Dec 4 '18 at 10:23
• Yes, that is the definition of continuity. – Julián Aguirre Dec 4 '18 at 10:34
• I was expecting a subtlety and I missed this obvious fact. I feel a bit fool about it, thank you for the refresh. – jlandercy Dec 4 '18 at 10:40