# Squeeze Theorem

> AP Calculus AB · Limits and Continuity (Unit 1)
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u1-squeeze-theorem/

This guide covers the formal statement, core conditions, and applications of the Squeeze Theorem for AP Calculus AB, including proofs of the fundamental trigonometric limit, evaluating limits at finite points and infinity, and exam-ready problem solving.

**Prerequisites:** Basic limit notation and the definition of a limit; Properties of inequalities for real numbers; [Evaluating limits via substitution and factoring](https://www.owlsprep.com/study/ap-calculus-ab-u1-evaluating-limits-analytically/)

## Learning objectives

- State the formal definition of the Squeeze Theorem
- Apply the Squeeze Theorem to evaluate limits of bounded oscillating functions
- Prove the fundamental trigonometric limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ using the Squeeze Theorem
- Apply the Squeeze Theorem to limits at infinity
- Avoid common mistakes when setting up inequalities for the Squeeze Theorem

## Definition and Core Conditions

The Squeeze Theorem (alternatively called the Sandwich Theorem or Pinching Theorem, names you may see on the AP exam) is a core tool for evaluating limits that cannot be simplified via factoring or substitution, most commonly oscillating bounded functions. Per the AP CED, it makes up ~2-3% of your total exam score, appearing mostly in multiple-choice but occasionally as a justification step in FRQs.

**Squeeze Theorem** — If $g(x) \leq f(x) \leq h(x)$ for all $x$ near $a$ (not required at $x=a$ itself), and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$. The theorem applies to all limit types: two-sided, one-sided, and limits at infinity.

Three core conditions must be satisfied for a valid application: (1) the inequality holds for all $x$ near the limit point (except possibly the limit point itself, which is irrelevant for limits), (2) the limits of both bounding functions exist and are equal, and (3) for limits at infinity, the inequality only needs to hold for all sufficiently large (or sufficiently negative) $x$. The most common initial challenge is finding valid bounds, which almost always starts from known bounds of sine and cosine.

**Worked example:** Verify all conditions of the Squeeze Theorem are satisfied to evaluate $\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)$

1. Start with the universal bound for sine, which holds for all $x \neq 0$:
2. $$-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$$
3. Multiply all parts of the inequality by $x^2$, which is non-negative for all real $x$, so inequality directions do not change. This gives the required bounds for our function:
4. $$-x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2$$
5. This satisfies the first condition, as it holds for all $x \neq 0$ near 0.
6. Compute the limits of the bounding functions:
7. $$\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0$$
8. Both limits exist and are equal, satisfying the second condition. All conditions are satisfied, so by the Squeeze Theorem:
9. $$\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0$$

> **Exam tip:** When multiplying an inequality by a function, always confirm the function is non-negative to avoid flipping inequality signs. Use absolute values to build bounds if the function can change sign.

## Proving the Fundamental Trigonometric Limit

One of the most important uses of the Squeeze Theorem on the AP exam is proving the fundamental trigonometric limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$, which is required to derive all derivative rules for trigonometric functions later in the course. The proof uses a geometric area argument for the unit circle, with $x$ measured in radians for $0 < x < \frac{\pi}{2}$, then extended to negative $x$ by symmetry.

- Area of inscribed triangle with angle $x$: $\frac{1}{2} \sin x$
- Area of circular sector with angle $x$: $\frac{1}{2} x$
- Area of circumscribed tangent triangle: $\frac{1}{2} \tan x$

$$\frac{1}{2} \sin x < \frac{1}{2} x < \frac{1}{2} \tan x$$

Multiply through by $\frac{2}{\sin x}$ (positive for $0 < x < \frac{\pi}{2}$, so no inequality flip):

$$1 < \frac{x}{\sin x} < \frac{1}{\cos x}$$

Take reciprocals of all positive terms, which flips the inequalities:

$$\cos x < \frac{\sin x}{x} < 1$$

**Worked example:** Use the Squeeze Theorem to prove $\lim_{x \to 0} \frac{\sin x}{x} = 1$ from the inequality above

1. The inequality $\cos x < \frac{\sin x}{x} < 1$ holds for all $x$ near 0 except $x=0$, satisfying the first condition of the Squeeze Theorem. Symmetry confirms it holds for both positive and negative $x$ near 0.
2. Evaluate the limit of the lower bound: by continuity of cosine at 0:
3. $$\lim_{x \to 0} \cos x = \cos 0 = 1$$
4. The upper bound is a constant function, so its limit is also 1:
5. $$\lim_{x \to 0} 1 = 1$$
6. Both bounds approach the same limit of 1, so by the Squeeze Theorem, the result follows:
7. $$\lim_{x \to 0} \frac{\sin x}{x} = 1$$

> **Exam tip:** This result only applies when the argument of sine and the denominator both approach 0. Never use it for limits approaching a non-zero value, where direct substitution works.

## Applications: Limits at Finite Points

The Squeeze Theorem is commonly used to evaluate limits of bounded oscillating functions multiplied by a term that approaches 0 at a finite point. Sine and cosine are always bounded between -1 and 1, so multiplying by a term that goes to 0 traps the product between two functions that both go to 0.

**Check your understanding**

Test your understanding of setup for finite limits:

1. What is $\lim_{x \to 0} x^4 \cos\left(\frac{e^x}{x^2}\right)$?

   - A) 0
   - B) 1
   - C) -1
   - D) The limit does not exist

   *Why:* Correct! We use the bound $-1 \leq \cos(\cdot) \leq 1$, multiply by non-negative $x^4$ to get $-x^4 \leq x^4 \cos(\cdot) \leq x^4$, both bounds go to 0, so the limit is 0.

**Worked example:** Let $f(x) = \frac{3x \sin^2 x}{1 + x^2}$ for all $x \neq 0$. (a) Use the Squeeze Theorem to find $\lim_{x \to 0} f(x)$ and justify your answer. (c) What value of $f(0)$ makes $f$ continuous at $x=0$?

1. Start with the known bound for squared sine: $0 \leq \sin^2 x \leq 1$ for all $x$. To avoid sign issues, we use absolute value to build bounds:
2. $$0 \leq \left| \frac{3x \sin^2 x}{1+x^2} \right| \leq \frac{3|x|}{1+x^2}$$
3. Expanding the absolute value inequality gives the final bounds:
4. $$-\frac{3|x|}{1+x^2} \leq f(x) \leq \frac{3|x|}{1+x^2}$$
5. Evaluate the limits of the bounds:
6. $$\lim_{x \to 0} -\frac{3|x|}{1+x^2} = 0 \quad \text{and} \quad \lim_{x \to 0} \frac{3|x|}{1+x^2} = 0$$
7. By the Squeeze Theorem, $\lim_{x \to 0} f(x) = 0$. For continuity at $x=0$, we require $\lim_{x \to 0} f(x) = f(0)$, so setting $f(0) = 0$ makes $f$ continuous at 0.

## Applications: Limits at Infinity

The Squeeze Theorem is also very commonly used for limits at infinity involving bounded oscillating functions. Any sine or cosine function remains bounded between -1 and 1 even as its input goes to infinity, so when multiplied by a term that approaches 0 as $x \to \pm \infty$, the entire product approaches 0 by the Squeeze Theorem. For limits at infinity, the only change to the conditions is that the inequality only needs to hold for all sufficiently large (or sufficiently negative) $x$, which is almost always satisfied for common problems.

**Worked example:** Evaluate $\lim_{x \to \infty} \frac{3\cos(2x^2) - 5}{x}$

1. Start with the universal bound for cosine, then manipulate step-by-step:
2. $$-1 \leq \cos(2x^2) \leq 1 \implies -3 \leq 3\cos(2x^2) \leq 3 \implies -8 \leq 3\cos(2x^2) - 5 \leq -2$$
3. For $x \to \infty$, $x$ is positive, so dividing by $x$ does not change inequality directions, giving us the bounds:
4. $$\frac{-8}{x} \leq \frac{3\cos(2x^2) - 5}{x} \leq \frac{-2}{x}$$
5. Compute the limits of the bounds:
6. $$\lim_{x \to \infty} \frac{-8}{x} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{-2}{x} = 0$$
7. Both bounds approach 0, so by the Squeeze Theorem:
8. $$\lim_{x \to \infty} \frac{3\cos(2x^2) - 5}{x} = 0$$

**Worked example:** A damped harmonic oscillator has vertical position (in cm) at time $t \geq 0$ given by $y(t) = \frac{5 \cos(8\pi t)}{t + 2}$. Use the Squeeze Theorem to find $\lim_{t \to \infty} y(t)$ and interpret the result.

1. Start with the universal bound for cosine, which holds for all $t$:
2. $$-1 \leq \cos(8\pi t) \leq 1$$
3. Multiply all parts by $\frac{5}{t+2}$, which is positive for all $t \geq 0$, so inequality directions do not change:
4. $$-\frac{5}{t+2} \leq y(t) \leq \frac{5}{t+2}$$
5. Evaluate the limits of the bounds:
6. $$\lim_{t \to \infty} -\frac{5}{t+2} = 0 \quad \text{and} \quad \lim_{t \to \infty} \frac{5}{t+2} = 0$$
7. By the Squeeze Theorem, $\lim_{t \to \infty} y(t) = 0$. In context, this means the oscillator's position approaches the equilibrium position of 0 cm over time, which matches the behavior of a damped oscillator losing energy.

> **Exam tip:** When dividing an inequality by $x$ for $x \to -\infty$, $x$ is negative, so you must flip the directions of the inequalities to get correct bounds.

## Common pitfalls

- **Wrong:** When evaluating $\lim_{x \to 0} x \sin(1/x)$, writing $-x \leq x \sin(1/x) \leq x$ without accounting for negative $x$, leading to reversed inequality directions for $x<0$.
  - Why it fails: Students forget multiplying by a negative number flips inequality signs, and do not adjust for the sign of $x$.
  - Correct: Use absolute value to build bounds: $|\sin(1/x)| \leq 1$ gives $|x \sin(1/x)| \leq |x|$, which expands to $-|x| \leq x \sin(1/x) \leq |x|$, which holds for all $x \neq 0$.
- **Wrong:** Applying the Squeeze Theorem with bounds that have different limits, for example concluding $\lim_{x \to 0} (x + \sin x) = 0$ because $x-1 \leq x+\sin x \leq x+1$, even though the bounds approach -1 and 1.
  - Why it fails: Students confuse squeezing with the general fact that the middle function lies between the bounds, even when the bounds do not share the same limit.
  - Correct: If the bounds have different limits, find tighter bounds or use a different technique like the sum law for limits.
- **Wrong:** Applying $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to $\lim_{x \to 4} \frac{\sin x}{x}$, concluding the limit is 1 instead of $\frac{\sin 4}{4}$.
  - Why it fails: Students memorize the formula but forget it only applies when the argument of sine and the denominator both approach 0.
  - Correct: Always check that the argument and denominator both approach 0 before applying this result; use direct substitution for limits at non-zero points.
- **Wrong:** Building incorrect bounds by writing $x^2 < x^2 \sin^2(1/x) < 2x^2$ for $\lim_{x \to 0} x^2 \sin^2(1/x)$, when $\sin^2(1/x)$ is between 0 and 1.
  - Why it fails: Students rush to build bounds without starting from the known bound of the oscillating function.
  - Correct: Always start from the known bound of the oscillating term (e.g., $0 \leq \sin^2(\text{anything}) \leq 1$) and build the inequality step-by-step.
- **Wrong:** For $x \to -\infty$, leaving inequality directions unchanged after dividing by negative $x$, leading to reversed bounds and a wrong conclusion.
  - Why it fails: Students assume $x$ is positive even when the limit goes to negative infinity.
  - Correct: Confirm the sign of $x$ before multiplying or dividing an inequality, and flip inequality signs whenever you multiply/divide by a negative value.

## Cheatsheet

| Category | Statement/Formula | Notes |
| --- | --- | --- |
| Squeeze Theorem (finite $a$) | If $g(x) \leq f(x) \leq h(x)$ near $a$, $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$ | Inequality does not need to hold exactly at $x=a$ |
| Squeeze Theorem (infinity) | If $g(x) \leq f(x) \leq h(x)$ for all $x > N$, $\lim_{x \to \infty} g(x) = \lim_{x \to \infty} h(x) = L$, then $\lim_{x \to \infty} f(x) = L$ | Same logic for $x \to -\infty$, inequality holds for $x < -N$ |
| Sine bound | $-1 \leq \sin(u(x)) \leq 1$ | Holds for any real input $u(x)$ |
| Cosine bound | $-1 \leq \cos(u(x)) \leq 1$ | Holds for any real input $u(x)$ |
| Squared trig bound | $0 \leq \sin^2(u(x)) \leq 1$, $0 \leq \cos^2(u(x)) \leq 1$ | Holds for any real input $u(x)$ |
| Fundamental trig limit | $\lim_{x \to 0} \frac{\sin x}{x} = 1$ | Only applies when argument and denominator both approach 0; angle in radians |
| Inequality rule | Multiplying/dividing by a negative reverses inequality direction | Always check the sign of the term you are multiplying/dividing by |
| Common result | $\lim (\text{bounded function}) \times (\text{term} \to 0) = 0$ | Most frequent Squeeze Theorem result for oscillating functions |

## What's next

Mastering the Squeeze Theorem is a critical prerequisite for working with all trigonometric limits, and it is required to justify core derivative rules for trigonometric functions later in Unit 2. The fundamental result $\lim_{x \to 0} \frac{\sin x}{x} = 1$, proven here via the Squeeze Theorem, is used directly to show that the derivative of $\sin x$ is $\cos x$, so understanding this derivation is necessary to earn full points for justification on AP FRQs. For AP Calculus AB, the Squeeze Theorem is most often tested on multiple-choice questions asking for limits of bounded oscillating functions, and it connects directly to checking continuity at removable discontinuities. Building comfort with setting up correct bounds for this theorem will save you time on exam day and help you avoid common limit errors.

- [Connecting multiple representations of limits](https://www.owlsprep.com/study/ap-calculus-ab-u1-connecting-multiple-representations-of-limits/)
- [Exploring types of discontinuities](https://www.owlsprep.com/study/ap-calculus-ab-u1-exploring-types-of-discontinuities/)
- [Defining continuity at a point](https://www.owlsprep.com/study/ap-calculus-ab-u1-defining-continuity-at-a-point/)

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