Answer
$2$
Work Step by Step
Let $f(x)=\int_{0}^{1} \dfrac{1}{\sqrt x} dx$
Now, $\lim\limits_{k \to 0^{}} f(x)= \lim\limits_{k \to 0} \int_{k}^{1} \dfrac{1}{\sqrt x} dx$
Thus, $\lim\limits_{k \to 0^{+}} [2 \sqrt x]_{k}^{1}=\lim\limits_{k \to 0} [2-2 \sqrt{(k)}]= 2-0=2$
