Squeeze Theorem

Problem 143: Calculate the following limit:

(1) $\begin{equation*} \lim_{x \to \infty} \frac{\cos(x)}{x^2}.\end{equation*}$

Solution: A little trick to do this problem is to remember that the absolute value of cosine is bounded by 1. That is,

(2) $\begin{align*} -1 \le \cos(x) \le 1 \quad \Rightarrow \quad -\frac{1}{x^2} \le \frac{\cos(x)}{x^2} \le \frac{1}{x^2}.\end{align*}$

Since $f(x)$ is bounded by two functions, says $g(x) = -\frac{1}{x^2}$ and $h(x) = \frac{1}{x^2}$ , we will use the Squeeze Theorem to find our limit.

The Squeeze Theorem: Suppose that $g(x) \le f(x) \le h(x)$ for all $x$ in some open interval containing $c$ , except possibly at $x = c$ itself. Suppose also that $\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L$ . Then $\lim_{x \to c} f(x) = L$ .

That is, since we are interested in the limit approaching $0$ which is in the interested interval,

(3) $\begin{align*}\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} = -\frac{1}{x^2} = 0\\ \lim_{x \to \infty} h(x) & = \lim_{x \to \infty} \frac{1}{x^2} =0. \end{align*}$

By the Squeeze Theorem,

(4) $\begin{equation*} \lim_{x \to \infty} f(x) = 0. \end{equation*}$

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