Single Variable Calculus, Chapter 4, 4.4, Section 4.4, Problem 30

Determine the $\displaystyle \lim_{x \to \infty} \sqrt{x} \sin \frac{1}{x}$



$
\begin{equation}
\begin{aligned}

\displaystyle \lim_{x \to \infty} \sqrt{x} \sin \frac{1}{x} \cdot \frac{\sqrt{x}}{\sqrt{x}} =& \lim_{x \to \infty} \frac{x}{\sqrt{x}} \sin \frac{1}{x} = \lim_{x \to \infty} \frac{1}{\sqrt{x}} x \sin \frac{1}{x}

&& \text{We can rewrite $\displaystyle \lim_{x \to \infty} x \sin \frac{1}{x}$ to $\displaystyle \lim_{x \to \infty} \frac{\displaystyle \sin \frac{1}{x}}{\displaystyle \frac{1}{x}}$}
\\
\\

=& \lim_{x \to \infty} \frac{1}{\sqrt{x}} \cdot \lim_{x \to \infty} \frac{\displaystyle \sin \frac{1}{x}}{\displaystyle \frac{1}{x}}
&&

\\
\\
=& (0)(1)
&&
\\
\\

=& 0
&&

\end{aligned}
\end{equation}
$


As $x \to \infty$, the leading term of a polynomial function will dominate the infinity, thus the other terms can be omitted.