Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 37

Prove that $\lim\limits_{x \rightarrow 0} x^4 \displaystyle \cos \frac{2}{x} = 0$


$\begin{equation} \begin{aligned} & \text{Proof:}\\ \phantom{x}& && \lim\limits_{x \to 0} x^4 \cos \frac{2}{x} = \lim\limits_{x \to 0} x^4 \cdot \lim\limits_{x \to 0} \cos \frac{2}{x}\\ \phantom{x}& && \lim\limits_{x \to 0} x^4 \cos \frac{2}{x} \text{ does not exist, the function is undefined because the denominator is equal to 0. However, since}\\ \phantom{x}& && - 1 \leq \cos \frac{2}{x} \leq 1\\ & \text{We have,}\\ \phantom{x}& && - x^4 \leq x^4 \cos \frac{2}{x} \leq x^4\\ & \text{We know that, }\\ \phantom{x}& && \lim\limits_{x \to 0} x^4 = -(0)^4 = 0 \text{ and } \lim\limits_{x \to 0} x^4 = (0)^4 = 0\\ & \text{Taking}\\ \phantom{x}& && f(x) = -x^4 \quad g(x) = x^4 \cos \frac{2}{x} \quad h(x) = x^4 \text{ in the squeeze theorem we obtain }\\ \phantom{x}& && \lim\limits_{x \to 0} x^4 \cos \frac{2}{x} = 0 \end{aligned} \end{equation}$