Answer
$0$
Work Step by Step
Notice that:
$$0 \leq \sin^{2}(x) \leq 1$$
$$0^{2} \leq (\sin^{2}(x))^{2} \leq 1^{2}$$
$$0 \leq \sin^{4}(x) \leq 1$$
$$0\cdot \frac{1}{\sqrt{x}} \leq \sin^{4}(x)\cdot \frac{1}{\sqrt{x}} \leq 1 \cdot \frac{1}{\sqrt{x}}$$
$$0\leq \frac{\sin^{4}(x)}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}$$
Since $\lim\limits_{x\to \infty}0=\lim\limits_{x\to \infty}\frac{1}{\sqrt{x}}=0$ then by the squeeze theorem it follows that:
$$\lim\limits_{x\to \infty}\frac{\sin^{4}(x)}{\sqrt{x}}=0$$
