I would like to calculate the following limit:

\begin{equation} A=\lim_{d\to 0^+}\exp\left[ -\left(\frac{d}{1-q}\right)\left(W_{0}\left[ B\left( 1+\frac{x}{rq}\right)^{\frac{1}{d}} \right]-W_{0}[B]\right)\right] \end{equation} where $q,B,r>0$, $x \in \mathbb{R}$ and $W_0$ is the $k=0$ branch of the Lambert-$W$ function defined as: \begin{equation} x=W(x)e^{W(x)} \end{equation} Mathematica says that the limit is equal to: \begin{equation} A=\left( 1+\frac{x}{rq} \right)^{\frac{1}{q-1}} \end{equation} which seems correct to me, since I do expect power laws.

But I would really like to be able to evaluate this limit on my own. I mean, Mathematica will not show me how to proceed in order to reach the result. I have made some attempts but this limit is a nasty one.

Is it possible that someone who has some experience with this kind of limits helps me out a bit?

Thank you!

  • $\begingroup$ Do you mean $d\to0^+$? $\endgroup$ – joriki Jun 26 '16 at 23:27
  • $\begingroup$ @joriki Yes that would be correct. I will fix that, thank you. $\endgroup$ – Mitscaype Jun 26 '16 at 23:29

The term $W_0[B]$ doesn't contribute, since the corresponding factor goes to $1$ with $d\to0$. For the other term, use

$$ W_0(x)=\log x-\log\log x+o(1) $$

(see Wikipedia). The $o(1)$ term doesn't contribute, so the limit is

$$ \begin{equation} \lim_{d\to 0^+}\exp\left[ -\left(\frac{d}{1-q}\right)W_{0}\left[ B\left( 1+\frac{x}{rq}\right)^{\frac{1}{d}} \right]\right]\\ = \lim_{d\to 0^+}\exp\left[ -\left(\frac{d}{1-q}\right)\left(\log\left(B\left( 1+\frac{x}{rq}\right)^\frac1d\right)-\log\log\left(B\left( 1+\frac{x}{rq}\right)^\frac1d\right)\right)\right]\\ = \lim_{d\to 0^+}\exp\left[-\left(\frac{d}{1-q}\right)\left(\log B+\frac1d\log\left(1+\frac{x}{rq}\right)-\log\left(\log B+\frac1d\log\left(1+\frac{x}{rq}\right)\right)\right)\right]\\ =\exp\left[\frac1{q-1}\log\left(1+\frac x{rq}\right)\right] \\ =\left(1+\frac x{rq}\right)^{\frac1{q-1}}\;. \end{equation} $$

  • 1
    $\begingroup$ I see. My attempts were really not as direct as this approach. Plus tbh I was not aware of this particular expansion. Thank you a lot for taking the time to help me with this one! $\endgroup$ – Mitscaype Jun 26 '16 at 23:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.