# Evaluate $\lim_{a \to0,Q \to\infty} \frac{Q}{2\pi}\big[ \log(z-a)-\log(z+a) \big]$

Can anyone help me find $$\lim_{a \to 0,Q \to\infty} \frac{Q}{2\pi}\left[ \log(z-a)-\log(z+a) \right]$$

Where $aQ=A$ where $A$ is kept constant. I know it is in the form $\mu/z$ for some $\mu$

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what is $z$? a real positive number? – Ilya Feb 2 '12 at 17:11
No, it's any complex number, this is part of a Fluid dynamics question, this function is a complex potential of a source and sink at $x=a,x=-a$ and $Q$ is the strength of the source. – Freeman Feb 2 '12 at 17:14

If you replace $Q$ by $A/a$ your expression equals

$$\frac{A}{2\pi } \lim_{a\to 0} \frac{1}{a} \log\big( \frac{z-a}{z+a} \big)$$

But since the functions $f(a) = \log\big( \frac{z-a}{z+a} \big)$ satisfies $f(0)=0$ this should remind you of something...

EDIT: Guess the way to go from here would be to write

$$\log\big( \frac{z-a}{z+a}\big) = (\log(z-a)-\log(z)) - (\log(z+a) - \log(z) )$$ and using that $$\lim_{a\to 0} \frac{1}{a}(\log(z+a) -\log(z) ) = \frac{1}{z}$$

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l'hopital's rule? – Freeman Feb 2 '12 at 17:20
Ah I see! That's clever. But what about the term with $log(z-a)$? – Freeman Feb 2 '12 at 17:46
Got it! Thanks so much! – Freeman Feb 2 '12 at 17:52

First, substitute $Q$ by $A/a$.

Then, using the formulae (1) $\exp(x) = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$, (2) $\log a-\log b = \log(a/b)$ and (3) $a\log x = \log x^a$, we find:

\begin{align*} \lim_{a\to 0} \log\left( \frac{z-a}{z+a}\right)^{1/a} &= \lim_{a\to 0}\log \left( 1 - \frac1{\frac{z+a}{2a}}\right)^{\frac{z+a}{2a}\frac{2}{z+a}} \\ &= \lim_{a\to 0} \frac{2}{z+a} \log \left( 1 - \frac1{\frac{z+a}{2a}}\right)^{\frac{z+a}{2a}} \\ &= \frac2{z} \log( e^{-1}) \\&= -\frac2{z} \end{align*}

Finally, multiply by $A/2\pi$. I found: $\mu = -A/\pi$.

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Are you sure this fits with the formula you stated for $e$? – Freeman Feb 2 '12 at 17:38
Okay, my formula for exp(x) was wrong, the result should fit. $\lim_{a\to 0}\log \left( 1 - \frac1{\frac{z+a}{2a}}\right)^{\frac{z+a}{2a}} = \lim_{n\to \infty} \log \left( 1 - \frac1{n}\right)^n = \exp(-1)$, because $\frac{z+a}{2a}\to\infty$ iff $a\to 0$. – Myself Feb 2 '12 at 17:51
Thank you for your help, but your result conflicts with testcase's, I've checked theirs, it seems to work! – Freeman Feb 2 '12 at 17:54
@LHS, the conflict is because I forgot to take the log :-) Does $\mu = -A/\pi$ agree with the above answer? (Which is, admitted, a much more standard approach.) – Myself Feb 2 '12 at 17:57
I see! Thank you so much though, this is a very nice way of doing it. – Freeman Feb 3 '12 at 21:40