Mathematical analysis

After noting that $I(x) \to 0$ as $x \to 0$ , we see that the given limit is of the indeterminate form $0/0$ , and so we can apply L'Hopital's rule. Now by the Fundamental Theorem of Calculus $\dfrac{d}{dx}I(x) = \sin(\sqrt{x})$ , so

$\large \displaystyle \lim_{x \to 0} \dfrac{3 I(x)}{2 x^{3/2}} = \lim_{x \to 0} \dfrac{3 \sin(\sqrt{x})}{3 \sqrt{x}} = \boxed{1}$ , where we have used the fact that $\displaystyle \lim_{u \to 0} \dfrac{\sin(u)}{u} = 1$ .