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:



$\begin{equation} \begin{aligned} \lim \limits_{x \to 0} \sqrt{x^3 + x^2}& \frac{\sin \pi}{x} = \lim \limits_{x \to 0} \sqrt{x^3 + x^2} \cdot \lim \limits_{x \to 0} \frac{\sin \pi}{x} \\ \lim \limits_{x \to 0} \sqrt{x^3 + x^2}& \frac{\sin \pi}{x} \text{ does not exist, the function is undefined because the denominator is equal to 0. However, since}\\ \phantom{x}& -1 \leq \sin \frac{\pi}{x} \leq 1\\ \text{We have, }\\ \phantom{x}& - \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2} \sin \frac{\pi}{x} \leq \sqrt{x^3+x^2}\\ \text{We know that, }\\ \phantom{x} & \lim \limits_{x \to 0} \sqrt{x^3 + x^2} = - \sqrt{0^3+0^2} = 0 \text{ and } \lim \limits_{x \to 0} \sqrt{x^3 + x^2}= \sqrt{0^3+0^2} = 0\\ \text{Taking} \\ \phantom{x} & f(x) = -\sqrt{x^3+x^2}, \quad g(x) = \sqrt{x^3+x^2} \sin \frac{\pi}{x}, \quad h(x) = \sqrt{x^3+x^2} \text{ in the squeeze theorem we obtain}\\ \phantom{x} & \lim \limits_{x \to 0} \sqrt{x^3 + x^2} \sin \frac{\pi}{x} = 0 \end{aligned} \end{equation}$