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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.
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.
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