# What is the Squeeze Theorem

$$\definecolor{importantColor}{RGB}{255,0,0} \definecolor{secondaryColor}{RGB}{255,0,255} \definecolor{tertiaryColor}{RGB}{0,102,51}$$

### Quick Overview

• If two functions squeeze together at a particular point, then any function trapped between them will get squeezed to that same point.
• The Squeeze Theorem deals with limit values, rather than function values.
• The Squeeze Theorem is sometimes called the Sandwich Theorem or the Pinch Theorem.

### Graphical Example

In the graph below, the lower and upper functions have the same limit value at $$x = a$$. The middle function has the same limit value because it is trapped between the two outer functions.

### Formal Statement

Suppose $$f(x) \leq g(x) \leq h(x)$$ for all $$x$$ in an open interval about $$a$$ (except possibly at $$a$$ itself). Further, suppose

$$\displaystyle\lim_{x\to a} f(x) = \lim_{x\to a} h(x) = L$$

Then, $$\displaystyle\lim\limits_{x\to a} g(x) = L$$

Note that the exception mentioned in the statement of the theorem is because we are dealing with limits. That means we're not looking at what happens at $$x = a$$, just what happens close by.

##### Example 1

Suppose there are three functions such that $$f(x)\leq g(x) \leq h(x)$$ when $$x$$ is near $$2$$.

Further, suppose $$f(x) =-\frac 1 3 x^3+x^2-\frac 7 3$$ and $$h(x) = \cos\left (\frac \pi 2 x\right)$$ (with $$x$$ measured in radians).

Determine $$\displaystyle\lim\limits_{x\to 2} g(x)$$

#### Solution

Step 1

Find $$\displaystyle\lim\limits_{x\to 2} f(x)$$

\begin{align*} \displaystyle\lim_{x\to 2} f(x) & = \lim_{x\to 2}\left(-\frac 1 3 x^3+x^2-\frac 7 3\right)\\[6pt] & = \left(-\frac 1 3 (2)^3+(2)^2-\frac 7 3\right)\\[6pt] & = \left(-\frac 8 3+4-\frac 7 3\right)\\[6pt] & = -1 \end{align*}

Step 2

Find $$\displaystyle\lim\limits_{x\to 2} h(x)$$

\begin{align*} \lim_{x\to 2} h(x) & = \lim_{x\to 2} \cos\left(\frac \pi 2 x\right)\\[6pt] & = \cos\left(\frac \pi 2 (2)\right)\\[6pt] & = \cos\left(\pi\right)\\[6pt] & = -1 \end{align*}

Step 3: Conclusion

Since $$f(x) \leq g(x) \leq h(x)$$ and $$\displaystyle\lim\limits_{x\to 2} f(x) = \displaystyle\lim\limits_{x\to 2} h(x) = -1$$, the Squeeze Theorem guarantees $$\displaystyle\lim\limits_{x\to 2} g(x) = -1$$ as well.

$$\displaystyle\lim\limits_{x\to 2} g(x) = -1$$

##### Example 2

Suppose there is a function, $$g(x)$$, such that $$f(x)\leq g(x)\leq h(x)$$ when $$x$$ is near $$-1$$.

Further, suppose $$f(x) =-\frac 1 4 x^2-\frac 1 2 x$$ and $$h(x) = \frac 1 3 x^2 + \frac 2 3 x + \frac 2 3$$

Determine $$\displaystyle\lim\limits_{x\to-1} g(x)$$.

#### Solution

Step 1

Determine $$\displaystyle\lim\limits_{x\to-1} f(x)$$

\\ \begin{align*} \displaystyle\lim_{x\to-1} f(x) & = \displaystyle\lim_{x\to-1} \left(-\frac 1 4 x^2-\frac 1 2 x\right)\\[6pt] & = -\frac 1 4(-1)^2-\frac 1 2(-1)\\[6pt] & = -\frac 1 4 + \frac 1 2\\[6pt] & = \frac 1 4 \end{align*} \\

Step 2

Determine $$\displaystyle\lim\limits_{x\to -1} h(x)$$

\\ \begin{align*} \displaystyle\lim_{x\to-1} h(x) & = \displaystyle\lim_{x\to-1} \left(\frac 1 3 x^2 + \frac 2 3 x + \frac 2 3\right)\\[6pt] & = \frac 1 3(-1)^2 + \frac 2 3(-1) + \frac 2 3\\[6pt] & = \frac 1 3 - \frac 2 3 + \frac 2 3\\[6pt] & = \frac 1 3 \end{align*} \\

Step 3: Conclusion

The two outer functions, $$f$$ and $$h$$, don't squeeze together (that is, their limits are different). Consequently, the most we can say about $$\displaystyle\lim\limits_{x\to-1} g(x)$$ is that it is somewhere between $$y = \frac 1 4$$ and $$y = \frac 1 3$$ if it exists at all.

$$\displaystyle\lim\limits_{x\to-1} g(x)$$ is indeterminate with the information provided.

An Important Limit: $$\displaystyle\lim\limits_{\theta \to 0} \frac {\sin \theta} \theta$$

The next few lessons will center around this and similar limits. The derivation shown below uses the Squeeze Theorem as well as some basic geometry and trigonometry.

#### Solution

Step 1

Some preliminary information you should recall.

• When angles are measured in radians, the length of a circular arc is $$s = r\theta$$ (link). On the unit-circle, this reduces to $$s = \theta$$.
• For points on the unit-circle, their $$y$$-coordinates are simply $$\sin\theta$$.
Step 2

Show that $$\sin \theta \leq \theta$$

In the image below, the length of the $${\color{secondaryColor}circular}$$ $${\color{secondaryColor}arc}$$, $${\color{secondaryColor}s = \theta}$$ is larger than the $${\color{importantColor}vertical}$$ $${\color{importantColor}line}$$ $${\color{importantColor}segment}$$, $${\color{importantColor}\sin \theta}$$ that reaches to the point on the circle.

This will be true for any angle $$\theta$$ since the arc must cover the same vertical distance as the line, but also extra horizontal distance as well.

Step 3

Show that $${\color{secondaryColor}\theta} \leq {\color{tertiaryColor}\tan \theta}$$

The same reasoning as in Step 1 leads to this conclusion as well. The $${\color{tertiaryColor}diagonal}$$ $${\color{tertiaryColor}line}$$ $${\color{tertiaryColor}segment}$$, $${\color{tertiaryColor}\tan \theta}$$ must cover the same vertical distance as the $${\color{secondaryColor}circular}$$ $${\color{secondaryColor}arc}$$, $${\color{secondaryColor}\theta}$$, but it also must cover a greater horizontal distance. To do this, it must be longer than the arc.

Step 4

Algebraically adjust the two inequalities so that we have $$\frac{\sin \theta} \theta$$ in the center.

\\ \begin{align*} {\color{importantColor}\sin \theta} & \leq {\color{secondaryColor}\theta} \leq {\color{tertiaryColor}\tan\theta}\\[6pt] {\color{importantColor}\sin \theta} & \leq {\color{secondaryColor}\theta} \leq {\color{tertiaryColor}\frac{\sin\theta}{\cos\theta}}\\[6pt] {\color{importantColor}\frac{\sin \theta}{\sin\theta}} & \leq {\color{secondaryColor}\frac{\theta}{\sin\theta}} \leq {\color{tertiaryColor}\frac{\sin\theta}{\cos\theta}\cdot \frac 1 {\sin\theta}}\\[6pt] {\color{importantColor}1} & \leq {\color{secondaryColor}\frac \theta {\sin \theta}} \leq {\color{tertiaryColor}\frac 1 {\cos \theta}}\\[6pt] {\color{importantColor}1} & \leq {\color{secondaryColor}\frac {\sin \theta}\theta} \leq {\color{tertiaryColor}\cos \theta}\\[6pt] \end{align*} \\

Step 5

Find the limit as $$x\to 0$$ for the two outer functions.

$${\color{importantColor}\displaystyle\lim_{\theta \to 0} 1} = {\color{tertiaryColor}\displaystyle\lim_{\theta \to 0} \cos \theta} = 1$$

Step 6

Use the Squeeze Theorem

Since $$y = \frac{\sin\theta}\theta$$ is trapped between $$y = 1$$ and $$y = \cos \theta$$, we can use the squeeze theorem to conclude that $$\displaystyle\lim\limits_{\theta\to0}\frac{\sin\theta}\theta = 1$$.

$$\displaystyle\lim\limits_{\theta\to 0} \frac{\sin \theta} \theta = 1$$