i'm having a bit of trouble with this problem, in all honesty it's the trig of this problem that's confusing me, i believe I've set the problem up correctly but any help would be greatly appreciated.
A source of strength $\Lambda$ is placed at the origin $(x,y) = (0,0)$ and an equal strength sink is placed on the positive y-axis $(x,y)=(0,a)$. Write the stream function $\psi(x,y)$ for this combination, then let $a \longrightarrow 0$ with $\lim_{a \longrightarrow 0} = \lambda$ and find the stream function again.
Below is my current conceptualization of the problem.
We know that the cartesian version of the streamfunction would be given by
$$\psi = \frac{m}{2\pi}\tan^{-1}\left(\frac{y}{x}\right)$$
but with the sink being directly over the y axis, then surely arctan is undefined and so im having problems figuring out how to start this...i've been asked to do this in Cartesian system and the only thing i can think of is defining it as the quotient of sin and cos
Note: let $R = \sqrt{a^2+x^2}~\&~R' = \sqrt{x^2+y^2}$
so i've gone with
$$\frac{sin B}{y+a}=\frac{sin(\frac{\pi}{2}-A)}{R}=\frac{cos A}{R} \implies \frac{(y+a)cos A}{R} = \frac{x(y+a)}{R'R}=sinB$$
then $cosB = \frac{x^2 + ay}{RR'}$
giving
$tanB = \frac{sinB}{cosB}=\frac{x(y+a)}{x^2+ay}$
then working the stream function we have
$$\psi = \lim_{a \longrightarrow 0} \frac{\Lambda}{2\pi}\left[\tan^{-1}\left(\frac{y}{x}\right) - \tan^{-1}\left( \frac{x(y+a)}{x^2+ay}\right)\right] \implies$$
$$\psi = \lim_{a \longrightarrow 0} \frac{\Lambda}{2\pi}\left[\tan^{-1}\left(\frac{\frac{y}{x}-\frac{x(y+a)}{x^2+ay}}{1+\frac{xy(y+a)}{x(x^2+ay)}}\right) \right] \implies$$
$$\frac{y}{x}-\frac{x(y+a)}{x^2+ay} = \frac{a(y^2-x^2)}{x(x^2+ay)}$$ and $$1+\frac{xy(y+a)}{x(x^2+ay)} = \frac{x(x^2+ay)+xy(y+a)}{x(x^2+ay)} =\frac{x(x^2+2ay+y^2)}{x(x^2+ay)}$$ then
$$\psi = \lim_{a \longrightarrow 0} \frac{\Lambda}{2\pi}\left[\tan^{-1}\left(\frac{a(y^2-x^2)}{x(x^2+2ay+y^2)}\right)\right] = \frac{\lambda}{2\pi}\frac{y^2 - x^2}{x(x^2+y^2)} \neq \frac{\lambda}{2\pi}\frac{x}{x^2+y^2}$$
which is apparently the solution im meant to have so....where have i gone wrong? any help would be greatly appreciated.
If a source/sink of strength $\Lambda$ placed at $(0,0)$, then its stream function is given by $$ \psi = -\frac{\Lambda}{2\pi}\tan^{-1}\frac yx. $$ Expression $\tan^{-1}\frac yx$ here is the angle of the point $(x,y)$, which makes sense since the streamlines of single source/sink are the straight lines going from the center. So it's natural to parametrize them as $\theta=\mathrm{const}$ in polar coordinates.
However, in Cartesian coordinates, it's impossible to parametrize this stream function smoothly over the whole region. At least at some line, there will be a discontinuity. For the expression $\tan^{-1}(y/x)$, the discontinuity happens at line $x=0$.
Your mistake is that you don't know what to do when sink/source isn't at the origin. However, it's easy to see that if sink/source is shifted to $(x_0,y_0)$, then it's stream function is merely: $$ \psi = -\frac{\Lambda}{2\pi}\tan^{-1}\frac {y-y_0}{x-x_0}. $$
Thus, the part with limit should be: $$ \lim_{a\to0}\Lambda\tan^{-1}\frac{\frac yx - \frac{y-a}{x}}{1+\frac{y(y-a)}{x^2}} = \frac{\lambda x}{x^2+y^2} $$
But the result can be also obtained from purely geometric observations:
The stream function is proportional to difference of angle arguments, which is geometrically is a small angle in the vertex of the triangle (angle $B$ on your picture). When $a$ is very small, this angle can be found as a ratio of normal segment of dipole $l$ (red) and the distance to the dipole $r$ (length $AB$ on your picture). $$ l = a\sin\alpha = a\frac{x}{\sqrt{x^2+y^2}},\qquad r = \sqrt{x^2+y^2},\\ \frac lr = \frac{a x}{x^2+y^2} $$