Find a limit for Doublet Stream function In fluid Mechanics,
The superimposed stream function of point source and sink is:
$\psi=-\frac{Qcos\theta_1}{4\pi}+\frac{Qcos\theta_2}{4\pi}$ 
Graphical image of the function
and for a sink - source doublet, we need to show that as:  $l\rightarrow 0$
we get:
$\psi=\frac{m}{r}sin^2\theta$
where: $m=\lim_{l\rightarrow0}\frac{Ql}{4\pi}$
I know it's a simple geometric limit problem , but somehow I can't derive the desired result.
Thanks!
 A: The positions of the sink and source are $(-l/2,0)$ and $(l/2,0),$ respectively.
Hence,
$$\psi = \frac{Q}{4 \pi}(\cos \theta_2 - \cos \theta_1) \\=  \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{(x-l/2)^2 + y^2}}- \frac{x+l/2}{\sqrt{(x+l/2)^2 + y^2}}\right) \\ =  \frac{Q}{4 \pi}\left( \frac{x-l/2}{\sqrt{x^2 + y^2 -lx +l^2/4}}- \frac{x+l/2}{\sqrt{x^2 + y^2 +lx + l^2/4}}\right) $$
In terms of polar coordinates, $r = \sqrt{x^2 +y^2}$  and
$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 - \frac{lx -l^2/4}{r^2}\right)^{-1/2}- \frac{x+l/2}{r}\left(1 + \frac{lx +l^2/4}{r^2}\right)^{-1/2}\right).$$
Using the Taylor expansion of the square root terms for small $l$ we obtain
$$\psi = \frac{Q}{4 \pi}\left( \frac{x-l/2}{r}\left(1 + \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)- \frac{x+l/2}{r}\left(1 - \frac{1}{2}\frac{lx}{r^2}+ O(l^2)\right)\right) \\ =\frac{Q}{4 \pi}\left(-\frac{l}{r}+ \frac{lx^2}{r^3} +O(l^2) \right) \\= \frac{Ql}{4 \pi r}\left(-1+ \frac{x^2}{r^2}  \right)+O(l^2),$$
and with $x = r\cos \theta,$
$$\psi = \frac{Ql}{4 \pi r}\left(-1+ \cos^2 \theta  \right)+O(l^2) \\ = -\frac{Ql}{4 \pi r}\sin^2 \theta+O(l^2).$$
Taking the limit as $l \to 0$ we get
$$\lim_{l \to 0} \psi = -\frac{m}{r}\sin^2 \theta$$
