# Some questions about the Lambert W function

1. Oct 23, 2011

### dimension10

I was wondering, what would be the solution for $$\lim_{x \rightarrow -\infty} W(x)$$ where W(x) is the Lambert-W function.

Wolfram|Alpha gives the result $$\infty$$ but the graph certainly does not imply this (In fact,$$\lim_{x\rightarrow \infty} x\,\exp(x)=\infty$$) I only graphed it it between -10 and 10 and behind -1.5 or so, the graph goes straight down so you can't even see the function beyond -1.75 or so. Does the function make a sharp turn somewhere else behind? Is the function even continuous?

Also,

$$\lim_{\omega \rightarrow -\infty} \omega \,\exp(\omega)=0$$ and $$\lim_{\omega\rightarrow 0} \omega\,\exp(\omega)=0$$. So does this imply that $$W(0)=0 \mbox{ and } W(-\infty)=0$$.

And one final question, Wolfram|Alpha gives a long complicated result for the inverse function of x^x involving the Lambert W function. How does one use the Lambert W function to do so?

Thanks.

Last edited: Oct 23, 2011
2. Oct 23, 2011

### Mute

For the real branches of the Lambert Function, $W_{0}(x)$ or $W_{-1}(x)$, the limit doesn't exist. The plot of x Exp[x] never dips down to -Infinity, so there is no limit. In fact, these two branches are only defined on the range $x \geq -1/e$. The complex branches may have a limit as x tends to -Infinity. That may be the result that wolframalpha is giving you.

As for x = y^y, write this as x = exp[y ln y] and then take the log of both sides. Then note that y = e^(ln y). (There are some other subtleties regarding branches of the complex logarithm, but this is the basic result).

3. Oct 24, 2011

Thanks