Answer
$$\lim_{x\to0^+}\frac{\sin x}{\sin\sqrt x}=0$$
Work Step by Step
$$A=\lim_{x\to0^+}\frac{\sin x}{\sin\sqrt x}$$
$$A=\lim_{x\to0^+}\frac{\sin x}{x}\times\lim_{x\to0^+}\frac{x}{\sin\sqrt x}$$
$$A=1\times\lim_{x\to0^+}\frac{x}{\sin\sqrt x}$$
$$A=\lim_{x\to0^+}\frac{\sqrt x\sqrt x}{\sin\sqrt x}$$
$$A=\lim_{x\to0^+}\frac{\sqrt x}{\sin\sqrt x}\times\lim_{x\to0^+}(\sqrt x)$$
$$A=\Big(\lim_{x\to0^+}\frac{\sin\sqrt x}{\sqrt x}\Big)^{-1}\times\sqrt0$$
As $x\ne0$, $\sqrt x\ne0$. Therefore, $\lim_{x\to0^+}\frac{\sin\sqrt x}{\sqrt x}=1$.
$$A=(1)^{-1}\times\sqrt0=1\times0=0$$
