Thomas' Calculus 13th Edition

$1$
The given function can be re-arranged as: $\lim\limits_{x \to \infty} \dfrac{\sqrt {x}}{\sqrt {\sin x}}=\sqrt{\lim\limits_{x \to 0^{+}} \dfrac{x}{\sin x}}$ and $\sqrt{ \dfrac{1}{\lim\limits_{x \to 0^{+}}(\dfrac{\sin x}{x})}}=\sqrt {\dfrac{1}{1}}=1$