Limit evaluation with elliptic integrals Prove the following involving elliptic integrals:
$$ \lim_{u\to 0 } \dfrac{K(u)- E(u) }  {1 - \sqrt {1-u}} = \frac{\pi}{2} $$

We have:
$$\frac{K(u)-E(u)}{1-\sqrt{1-u}}=\frac{u}{1-\sqrt{1-u}}\int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1-u\sin^2\theta}}\,d\theta$$
and since:
$$\lim_{u\to 0}\frac{u}{1-\sqrt{1-u}}=2$$
we only have to notice that:
$$\lim_{u\to 0}\int_{0}^{\pi/2}\frac{\sin^2\theta}{\sqrt{1-u\sin^2\theta}}\,d\theta=\int_{0}^{\pi/2}\sin^2\theta\,d\theta = \frac{\pi}{4}$$
by the dominated convergence theorem.
