Prove or disprove: $\lim_{a\to 1^-}E(a)=1$ for $E(a)$ elliptic integral of second kind

Question

Let $E(a)=\int\limits_0^{\frac{\pi}{2}}\sqrt{1-a^2\sin^2t}\,\mathrm{d}t,~~0<a<1$ be the elliptic integral of second kind. Prove or disprove: $$\displaystyle \lim_{a\to 1^-}E(a)=\int\limits_0^{\frac{\pi}{2}}\sqrt{1-\sin^2t}\,\mathrm{d}t=1.$$

Attempt.
$$0<E(a)-\int\limits_{0}^{\frac{\pi}{2}}\sqrt{1-\sin^2x}\,\mathrm{d}x$$ $$=\left(1-a^2\right) \int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-a^2 \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$$ so we need to evaluate the limit of the last expression (in my post Existence of $\lim_{k\to +\infty}\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^2x}{\sqrt{1-\frac{k^2}{k^2+1} \,\sin^2x} + \sqrt{1-\sin^2x}}\,\mathrm{d}x$ the answer was that the sequence tends to $+\infty$, so in our case we get indefinite form $0\cdot\infty$).

Thanks for the help.

Answer 1

The integrand $f(t, a) $ is continuous on $[0,\pi/2]\times[0,1]$ and hence the integral $E(a) $ is continuous on $[0,1]$. It follows trivially that $$\lim_{a\to 1^{-}}E(a)=E(1)=1$$

Answer 2

Using Mathematica notations, $$\int\limits_0^{\frac{\pi}{2}}\sqrt{1-a^2\sin^2t}\,dt=E\left(a^2\right)$$ and its expansion around $a=1$ is given by $$E\left(a^2\right)\sim 1-\frac{1}{2} (1-a) \left(\log \left(\frac{1-a}{8}\right)+1\right)+\cdots$$

Let $a=1-10^{-k}$ and compare the numerical values $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 1.16910133173369 & 1.17169705278161 \\ 2 & 1.02842305863834 & 1.02847580902880 \\ 3 & 1.00399359841033 & 1.00399440996551 \\ 4 & 1.00051448909568 & 1.00051450008378 \\ 5 & 1.00006296183503 & 1.00006296197369 \\ 6 & 1.00000744747605 & 1.00000744747772 \end{array} \right)$$