How to find the limit of irrotational flow past a finite plate?

Question

In chapter 4 of Acheson's Fluid Dynamics we have the equation $$u_*-iv_*=\left(Ue^{-i\alpha}-Ue^{i\alpha}\frac{a^2}{(a+\varepsilon)^2}-\frac{i\Gamma}{2\pi (a+\varepsilon)}\right)\bigg/\left(1-\frac{a^2}{(a+\varepsilon)^2}\right)$$ which is the same as $$u_*-iv_*=\left(Ue^{-i\alpha}-Ue^{i\alpha}\frac{a^2}{(a+\varepsilon)^2}+\frac{i4\pi Ua\sin\alpha}{2\pi (a+\varepsilon)}\right)\bigg/\left(1-\frac{a^2}{(a+\varepsilon)^2}\right)$$ on account of value found for $\Gamma$. While for $\varepsilon=0$ both numerator and denominator vanish (necessary to remove the singularities), the author states that as $\varepsilon\to0$, $u_*$ goes to $U\cos\alpha$ while the imaginary part vanishes, and I don't see how he finds this.

Answer 1

It's simple algebra. The real part of the numerator is $$ U\cos\alpha-U\cos\alpha\,\frac{a^2}{(a+\varepsilon)^2} = U\cos\alpha\left(1-\frac{a^2}{(a+\varepsilon)^2}\right), $$ and its imaginary part is \begin{align} -U\sin\alpha-U\sin\alpha\,\frac{a^2}{(a+\varepsilon)^2} + \frac{2Ua\sin\alpha}{a+\varepsilon} &= -U\sin\alpha\left(1-\frac{2a}{a+\varepsilon}+\frac{a^2}{(a+\varepsilon)^2}\right) \\ &=-U\sin\alpha\left(1-\frac{a}{a+\varepsilon}\right)^2. \end{align} Dividing by the denominator and taking the limit $\varepsilon\to 0$ we finally obtain $$ u_* = \lim_{\varepsilon\to 0} \frac{U\cos\alpha\left(1-\frac{a^2}{(a+\varepsilon)^2}\right)}{\left(1-\frac{a^2}{(a+\varepsilon)^2}\right)} = U\cos\alpha $$ and $$ v_* = -\lim_{\varepsilon\to 0} \frac{U\sin\alpha\,\left(1-\frac{a}{a+\varepsilon}\right)^2} {\left(1-\frac{a^2}{(a+\varepsilon)^2}\right)} = -\lim_{\varepsilon\to 0} \frac{U\sin\alpha\left(1-\frac{a}{a+\varepsilon}\right)} {\left(1+\frac{a}{a+\varepsilon}\right)} = 0. $$