Squeeze theorem — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Formal statement of the squeeze theorem, using the theorem to evaluate limits of trigonometric functions, limits at removable discontinuities, and limits at infinity, including derivation of the standard ${\lim }_{x\to 0}\frac{\sin x}{x}$ result.

You should already know: Basic limit notation and limit definition. Properties of inequalities for real numbers. Evaluating limits via substitution and factoring.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Squeeze theorem?

The Squeeze Theorem (also called the Sandwich Theorem or Pinching Theorem, alternate names it may use on the AP exam) is a core limit evaluation tool in Unit 1: Limits and Continuity. Per the AP Calculus AB CED, it accounts for approximately 2-3% of total exam score, appearing most commonly in multiple-choice questions (MCQ), but occasionally as a required justification step in longer free-response questions (FRQ) involving trigonometric limits or continuity.

Formal definition: If we can find two functions $g(x)$ (lower bound) and $h(x)$ (upper bound) such that $g(x)\le f(x)\le h(x)$ for all $x$ near $a$ (we do not need the inequality to hold exactly at $x=a$, per the definition of a limit), and ${\lim }_{x\to a}g(x)={\lim }_{x\to a}h(x)=L$, then ${\lim }_{x\to a}f(x)=L$ as well. The intuition is that $f(x)$ is squeezed between two functions that both approach the same value $L$, so $f(x)$ has no choice but to also approach $L$. It is especially useful for oscillating functions that cannot be factored or simplified directly, since their bounded oscillation can be trapped between two simple functions with matching limits.

2. Core Conditions and Setup for the Squeeze Theorem

The Squeeze Theorem only produces a valid result if three core conditions are met, and misapplying any condition leads to an incorrect answer. First, the inequality $g(x)\le f(x)\le h(x)$ must hold for all $x$ in an open interval around the limit point $a$, except possibly at $x=a$ itself. A violation of the inequality exactly at $x=a$ is irrelevant, because limits do not depend on the value of the function at the limit point. Second, the limits of both $g(x)$ and $h(x)$ as $x$ approaches $a$ must exist and be equal. If the two limits are different, or one does not exist, the theorem gives no conclusion about ${\lim }_{x\to a}f(x)$. Third, the theorem works for all limit types: two-sided $x\to a$, one-sided $x\to a^{+}$ or $x\to a^{-}$, and limits at infinity $x\to \pm \infty$ — the only change is adjusting where the inequality must hold (for $x\to \infty$, the inequality only needs to hold for all $x$ greater than some large number $N$).

The most common initial challenge is finding valid bounds that meet the conditions, which relies on starting from known bounds for common oscillating functions like sine and cosine.

Worked Example

Problem: Verify that all conditions of the Squeeze Theorem are satisfied to evaluate ${\lim }_{x\to 0}{x}^2\sin \left(\frac{1}{x}\right)$.

1. We start with the universal bound for sine: $-1\le \sin \left(\frac{1}{x}\right)\le 1$ for all $x\ne 0$.
2. Multiply all parts of the inequality by $x^2$, which is non-negative for all real $x$, so inequality directions do not change: $-x^2\le x^2\sin \left(\frac{1}{x}\right)\le x^2$. This holds for all $x\ne 0$, so it holds on any open interval around 0 (except $x=0$, which is allowed), satisfying the first condition.
3. Compute the limits of the bounds: ${\lim }_{x\to 0}(-x^2)=0$ and ${\lim }_{x\to 0}{x}^2=0$. Both limits exist and are equal, satisfying the second condition.
4. All conditions are satisfied, so by the Squeeze Theorem, ${\lim }_{x\to 0}{x}^2\sin \left(\frac{1}{x}\right)=0$.

Exam tip: When multiplying an inequality by a function, always confirm the function is non-negative to avoid flipping inequality signs; if the function can be negative, use absolute value to build bounds instead of splitting into cases.

3. Deriving the Fundamental Trigonometric Limit ${\lim }_{x\to 0}\frac{\sin x}{x}$

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 relies on a geometric argument for the unit circle, with $x$ measured in radians, for $0<x<\frac{\pi }{2}$ (the result extends to negative $x$ by symmetry).

For an angle $x$ in the first quadrant of the unit circle:

• The area of the inscribed triangle with base 1 and height $\sin x$ is $\frac{1}{2}\sin x$
• The area of the circular sector with angle $x$ is $\frac{1}{2}x$
• The area of the circumscribed tangent triangle with base 1 and height $\tan x$ is $\frac{1}{2}\tan x$

Comparing these areas gives the inequality: $$\frac{1}{2}\sin x<\frac{1}{2}x<\frac{1}{2}\tan x$$ Multiply through by $\frac{2}{\sin x}$ (positive for $0<x<\frac{\pi }{2}$, so inequalities do not flip): $$1<\frac{x}{\sin x}<\frac{1}{\cos x}$$ Take reciprocals of all positive terms, which flips the inequalities: $$\cos x<\frac{\sin x}{x}<1$$ This inequality holds for all $x$ near 0 except $x=0$, so we can apply the Squeeze Theorem.

Worked Example

Problem: Use the inequality $\cos x<\frac{\sin x}{x}<1$ and the Squeeze Theorem to prove ${\lim }_{x\to 0}\frac{\sin x}{x}=1$.

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: ${\lim }_{x\to 0}\cos x=\cos 0=1$, by continuity of cosine at 0.
3. The upper bound is the constant function 1, so ${\lim }_{x\to 0}1=1$.
4. Both bounds have the same limit 1, so by the Squeeze Theorem, ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, completing the proof.

Exam tip: This limit only applies when the argument of sine and the denominator both approach 0. Do not use it for limits approaching a non-zero value, where direct substitution works instead.

4. Using Squeeze Theorem for Limits at Infinity

The Squeeze Theorem is also commonly used to evaluate limits at infinity for bounded oscillating functions multiplied by a term that approaches 0. Any sine or cosine function is bounded between -1 and 1 for all real inputs, even when the input goes to infinity, so when multiplied by a term that approaches 0 as $x\to \pm \infty$, the entire product will approach 0 by the Squeeze Theorem.

For limits at infinity, the only requirement for the inequality is that it holds for all sufficiently large (or sufficiently negative, for $x\to -\infty$) $x$, which is easy to satisfy for common bounded functions. The logic of the theorem is identical to limits at finite points.

Worked Example

Problem: Evaluate ${\lim }_{x\to \infty }\frac{3\cos (2x^2)-5}{x}$.

1. Start with the universal bound for cosine: $-1\le \cos (2x^2)\le 1$ for all $x$. Multiply by 3: $-3\le 3\cos (2x^2)\le 3$. Subtract 5: $-8\le 3\cos (2x^2)-5\le -2$.
2. Divide all parts by $x$, which is positive for $x\to \infty$, so inequality directions stay the same: $\frac{-8}{x}\le \frac{3\cos (2x^2)-5}{x}\le \frac{-2}{x}$. This holds for all $x>0$, which satisfies the requirement for a limit as $x\to \infty$.
3. Compute the limits of the bounds: ${\lim }_{x\to \infty }\frac{-8}{x}=0$ and ${\lim }_{x\to \infty }\frac{-2}{x}=0$. Both limits exist and are equal.
4. By the Squeeze Theorem, ${\lim }_{x\to \infty }\frac{3\cos (2x^2)-5}{x}=0$.

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

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

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

Worked Solution: We start with the universal bound for cosine: $-1\le \cos \left(\frac{e^x}{x^2}\right)\le 1$ for all $x\ne 0$. Multiply all terms by $x^4$, which is non-negative for all real $x$, so inequality directions do not change: $-x^4\le x^4\cos \left(\frac{e^x}{x^2}\right)\le x^4$. Next, compute the limits of the bounds: ${\lim }_{x\to 0}(-x^4)=0$ and ${\lim }_{x\to 0}{x}^4=0$. Both bounds approach the same limit 0, so by the Squeeze Theorem, the limit of the middle function is also 0. Correct answer: A.

Question 2 (Free Response)

Let $f(x)=\frac{3x{\sin }^{2}x}{1+x^2}$ for all real $x\ne 0$. (a) Use the Squeeze Theorem to find ${\lim }_{x\to 0}f(x)$. Justify your answer. (b) Use the Squeeze Theorem to find ${\lim }_{x\to \infty }f(x)$. Justify your answer. (c) What value of $f(0)$ makes $f$ continuous at $x=0$? Explain.

Worked Solution: (a) We know $0\le {\sin }^{2}x\le 1$ for all $x\ne 0$. Multiply all terms by $\frac{3|x|}{1+x^2}$ (non-negative for all $x$) to get: $$0\le \left|\frac{3x{\sin }^{2}x}{1+x^2}\right|⟹-\frac{3|x|}{1+x^2}\le f(x)\le \frac{3|x|}{1+x^2}$$ Evaluate the bounds: ${\lim }_{x\to 0}-\frac{3|x|}{1+x^2}=0$ and ${\lim }_{x\to 0}\frac{3|x|}{1+x^2}=0$. By Squeeze Theorem, ${\lim }_{x\to 0}f(x)=0$.

(b) For $x>0$, $\frac{3x}{1+x^2}$ is positive, so we use the bound $0\le {\sin }^{2}x\le 1$ to get: $$0\le \frac{3x{\sin }^{2}x}{1+x^2}\le \frac{3x}{1+x^2}$$ Evaluate the bounds: ${\lim }_{x\to \infty }0=0$, and ${\lim }_{x\to \infty }\frac{3x}{1+x^2}={\lim }_{x\to \infty }\frac{3/x}{1/x^2+1}=0$. By Squeeze Theorem, ${\lim }_{x\to \infty }f(x)=0$.

(c) A function is continuous at $x=a$ if ${\lim }_{x\to a}f(x)=f(a)$. From part (a), ${\lim }_{x\to 0}f(x)=0$, so setting $f(0)=0$ makes $f$ continuous at $x=0$.

Question 3 (Application / Real-World Style)

A damped harmonic oscillator on a spring has its vertical position (in centimeters) at time $t\ge 0$ seconds given by $y(t)=\frac{5\cos (8\pi t)}{t+2}$, where $y=0$ is the equilibrium position of the oscillator. Use the Squeeze Theorem to find ${\lim }_{t\to \infty }y(t)$. Interpret your result in the context of the problem.

Worked Solution: We know for all real $t$, $-1\le \cos (8\pi t)\le 1$. Multiply all parts by $\frac{5}{t+2}$, which is positive for all $t\ge 0$, so inequality directions stay the same: $$-\frac{5}{t+2}\le y(t)\le \frac{5}{t+2}$$ Compute the limits of the bounds: ${\lim }_{t\to \infty }-\frac{5}{t+2}=0$ and ${\lim }_{t\to \infty }\frac{5}{t+2}=0$. By the Squeeze Theorem, ${\lim }_{t\to \infty }y(t)=0$. In context, this means that as time increases, the oscillator's position approaches the equilibrium position of 0 cm, which matches the expected behavior of a damped oscillator that loses energy over time.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the Squeeze Theorem is a prerequisite for evaluating all trigonometric limits, which come up repeatedly in the definition of the derivative later in Unit 2, when you derive derivative rules for sine and cosine. The core result ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, proved via the Squeeze Theorem, is used directly to show that the derivative of $\sin x$ is $\cos x$, so without understanding this derivation, you will not be able to justify these core derivative rules on FRQ questions. For AP Calculus AB, the Squeeze Theorem is most often tested on MCQ questions asking for limits of bounded oscillating functions, and it connects directly to the definition of continuity for functions with removable discontinuities. The follow-on topics that build on this chapter are: Derivatives of trigonometric functions Continuity at a point Evaluating limits analytically Limits at infinity