Answer
$\dfrac{\pi}{2}$
Work Step by Step
Let $f(x)=\int_{0}^{1} \dfrac{1}{\sqrt{1-x^2}} dx$
Consider $x=\sin p $ and $ dx=\cos p dp$
This implies that, $\int_{0}^{k} \dfrac{1}{\sqrt{1-x^2}} dx=\int_{0}^{\sin ^{-1} (k)} \dfrac{\cos p}{\sqrt{1-\sin^2 p}} dp$
Now, $\lim\limits_{k \to 1^{-}} \int_{0}^{\sin ^{-1} (k)} \dfrac{\cos (p)}{\sqrt{1-\sin^2 (p)}} dp=\lim\limits_{k \to 1^{-}}\sin^{-1} (k)$
Thus, $\lim\limits_{k \to 1^{-}}\sin^{-1} (k) =\sin^{-1} (\sin \dfrac{\pi}{2}) =\dfrac{\pi}{2}$
