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 $\mu$
Thanks very much in advance
 A: 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 
$$ \lim_{a\to 0} \frac{1}{a}(\log(z+a) -\log(z) ) = \frac{1}{z}$$
A: 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: $\mu = -A/\pi$. 
