Chapter 3 - Applications of Differentiation - 3.4 Limits at Infinity; Horizontal Asymptotes - 3.4 Exercises - Page 242: 32

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Given $$\lim _{x \rightarrow \infty}\sqrt{x}\sin\frac{1}{x}$$ Then \begin{aligned} \lim _{x \rightarrow \infty}\sqrt{x}\sin\frac{1}{x}&=\lim _{x \rightarrow \infty}\sqrt{x}\sin\left(\frac{1}{x}\right)\frac{1/x}{1/x} \\ &=\lim _{x \rightarrow \infty}\frac{1}{\sqrt{x}}\frac{\sin\left(\frac{1}{x}\right)}{1/x}\\ &=\lim _{x \rightarrow \infty}\frac{1}{\sqrt{x}}\lim _{x \rightarrow \infty}\frac{\sin\left(\frac{1}{x}\right)}{1/x}\\ &=(0)(1)\\&=0 \end{aligned}