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, $\lim_{x \to \infty} \frac{\sin x}{x}$.

We first note an important property of the sine function that is $-1 ≤ \sin x ≤ 1$. If we multiply all terms in this equality by [[$ \frac{1}{x}, we thus get that:

(4)

\begin{align} -\frac{1}{x} ≤ \frac{\sin x}{x} ≤ \frac{1}{x} \end{align}

We note that $\lim_{x \to \infty} -\frac{1}{x} = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$. Therefore it follows by the Squeeze theorem that since $-\frac{1}{x} ≤ \frac{\sin x}{x} ≤ \frac{1}{x}$, then $\lim_{x \to \infty} \frac{\sin x}{x} = 0$.

Example 2

Evaluate the following limit, $\lim_{x \to 0} x^2 \sin (\frac{1}{x})$.

Once again, we recall that $-1 ≤ \sin x ≤ 1$. It also follows that $-1 ≤ \sin (\frac{1}{x}) ≤ 1$. If we multiply both sides of this inequality by $x^2$ we get that:

(5)

\begin{align} -x^2 ≤ x^2 \sin (\frac{1}{x}) ≤ x^2 \end{align}

But $\lim_{x \to 0} -x^2 = 0$ and $\lim_{x \to 0} x^2 = 0$. Therefore by the Squeeze theorem, $\lim_{x \to 0} x^2 \sin (\frac{1}{x}) = 0$.

Example 3

Evaluate the following limit, $\lim_{x \to -\infty} \frac{5x^2 - \sin (3x)}{x^2 + 10}$.

Once again we recall that $-1 ≤ - \sin (3x) ≤ 1$. If we add $5x^2$ to each part of the inequality we get that $5x^2 - 1 ≤ 5x^2 - \sin (3x) ≤ 5x^2 + 1$. Lastly, we will divide each side of this inequality by $x^2 + 10$ to get the following inequality:

(6)

\begin{align} \frac{5x^2 - 1}{x^2 + 10} ≤ \frac{5x^2 - \sin (3x)}{x^2 + 10} ≤ \frac{5x^2 + 1}{x^2 + 10} \end{align}

We now compute $\lim_{x \to -\infty} \frac{5x^2 - 1}{x^2 + 10}$ as follows:

(7)

\begin{align} \lim_{x \to -\infty} \frac{5x^2 - 1}{x^2 + 10} = \lim_{x \to -\infty} \frac{5x^2 - 1}{x^2 + 10} \cdot \frac{1/x^2}{1/x^2} \\ \lim_{x \to -\infty} \frac{5x^2 - 1}{x^2 + 10} = \lim_{x \to -\infty} \frac{5 - \frac{1}{x^2}}{1 + \frac{10}{x^2}} = 5 \end{align}

We will now compute the other limit, $\lim_{x \to -\infty} \frac{5x^2 + 1}{x^2 + 10}$:

(8)

\begin{align} \lim_{x \to -\infty} \frac{5x^2 + 1}{x^2 + 10} = \lim_{x \to -\infty} \frac{5x^2 + 1}{x^2 + 10} \cdot \frac{1/x^2}{1/x^2} \\ \lim_{x \to -\infty} \frac{5x^2 + 1}{x^2 + 10} = \lim_{x \to -\infty} \frac{5 + \frac{1}{x^2}}{1 + \frac{10}{x^2}} = 5 \end{align}

By the Squeeze theorem, since $\frac{5x^2 - 1}{x^2 + 10} ≤ \frac{5x^2 - \sin (3x)}{x^2 + 10} ≤ \frac{5x^2 + 1}{x^2 + 10}$, it follows that $\lim_{x \to -\infty} \frac{5x^2 - \sin (3x)}{x^2 + 10} = 5$.