Squeeze Theorem

# Squeeze Theorem

 Theorem 1 (Squeeze Theorem): Suppose that $f$, $h$, and $g$ are functions such that $f(x) ≤ g(x) ≤ h(x)$ when $x$ is near $a$, though not necessarily at $a$. If $\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} g(x) = L$ as well.
• Proof: Suppose that $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} h(x) = L$. By the definition of a limit, it thus follows that for some $\delta_1, \delta_2, > 0$ and $\epsilon > 0$:
(1)
\begin{align} \quad \mathrm{if} \quad 0 < \: \mid x - a \mid \: < \delta_1 \quad \mathrm{then} \quad L - \epsilon \: < \: f(x) \: < \: L + \epsilon \\ \end{align}
• Similarly,
(2)
\begin{align} \quad \mathrm{if} \quad 0 < \: \mid x - a \mid \: < \delta_2 \quad \mathrm{then} \quad L - \epsilon \: < \: f(x) \: < \: L + \epsilon \\ \end{align}
• We will now let $S =\mathrm{min}(\delta_1, \delta_2)$. We note that if $S =\mathrm{min}(\delta_1, \delta_2)$, then $0 < \: \mid x - a \mid \: < \delta_1$ AND $0 < \: \mid x - a \mid \: < \delta_2$ since we recall that all values $p ≤ \delta$ are sufficiently small. Clearly $\mathrm{min}(\delta_1, \delta_2) ≤ \delta_1$ and $\mathrm{min}(\delta_1, \delta_2) ≤ \delta_2$, and therefore:
(3)
\begin{align} \quad L - \epsilon < f(x) ≤ g(x) ≤ h(x) < L + \epsilon \\ L - \epsilon < g(x) < L + \epsilon \end{align}
• This implies that $\mid g(x) - L \mid < \epsilon$ and therefore, $\lim_{x \to a} g(x) = L$ too. $\blacksquare$

We will now look at some examples applying the squeeze theorem.

## Example 1

Evaluate the following limit, $\lim_{x \to \infty} \frac{\sin x}{x}$.