Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 34

Graph the functions $f(x) = - \sqrt{x^3 + x^2}, g(x) = \sqrt{x^3 + x^2}$ and $h(x) = \sqrt{x^3 + x^2}\displaystyle \sin \frac{\pi}{x}$
on the same screen and using squeeze theorem, show that $\lim \limits_{x \to 0} \sqrt{x^3 + x^2} \displaystyle \sin \frac{\pi}{x} = 0$








Proof:



$
\begin{equation}
\begin{aligned}
\lim \limits_{x \to 0} \sqrt{x^3 + x^2}& \frac{\sin \pi}{x} = \lim \limits_{x \to 0} \sqrt{x^3 + x^2} \cdot \lim \limits_{x \to 0} \frac{\sin \pi}{x} \\
\lim \limits_{x \to 0} \sqrt{x^3 + x^2}& \frac{\sin \pi}{x} \text{ does not exist, the function is undefined because the denominator is equal to 0. However, since}\\
\phantom{x}& -1 \leq \sin \frac{\pi}{x} \leq 1\\

\text{We have, }\\

\phantom{x}& - \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2} \sin \frac{\pi}{x} \leq \sqrt{x^3+x^2}\\

\text{We know that, }\\

\phantom{x} & \lim \limits_{x \to 0} \sqrt{x^3 + x^2} = - \sqrt{0^3+0^2} = 0 \text{ and } \lim \limits_{x \to 0} \sqrt{x^3 + x^2}= \sqrt{0^3+0^2} = 0\\

\text{Taking} \\
\phantom{x} & f(x) = -\sqrt{x^3+x^2}, \quad g(x) = \sqrt{x^3+x^2} \sin \frac{\pi}{x}, \quad h(x) = \sqrt{x^3+x^2} \text{ in the squeeze theorem we obtain}\\

\phantom{x} & \lim \limits_{x \to 0} \sqrt{x^3 + x^2} \sin \frac{\pi}{x} = 0
\end{aligned}
\end{equation}
$