Asymptotic expansion of the complete elliptic integral of the first kind

You can employ the substitution $y=1- k^2 \sin^2x$ such that the integral can be written as $$ K(k) = \int_{1-k^2}^1\!dy\,\frac{1}{2 \sqrt{y(1-y)(y-1 +k^2)}}.$$

Now, we can expand the integrand in $\epsilon = 1-k$ and obtain $$ K(k) = \int_{1-k^2}^1\!dy\frac{1}{2 y \sqrt{1-y}} + O(1). \tag{1}$$ The estimate of the error term follows from the fact that expanding in $\eta =1-k^2$, we have $$K(k) = \sum_{n=0}^\infty c_n \eta^n \int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2}$$ with $c_n$ some constants. Now, we have for $n>0$ that $$ \int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2} = O(\eta^{-n})$$ such that only the $n=0$ term diverges for $\eta \to 0$ (which corresponds to $\epsilon \to 0$).

It thus remains to estimate the first term in (1) for $k \to 1$ which is not that difficult:

In fact due to the $1/y$ behavior close to $y=0$, the integral is logarithmically divergent and we have that $$K(k) = \frac12 \left|\log (1-k^2)\right| + O(1) = \frac12 \left|\log \epsilon\right| + O(1).$$