Squeeze Theorem — AP Calculus AB
1. Definition and Core Conditions ★★☆☆☆ ⏱ 3 min
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.
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.
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.
2. Proving the Fundamental Trigonometric Limit ★★★☆☆ ⏱ 4 min
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
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.
3. Applications: Limits at Finite Points ★★☆☆☆ ⏱ 3 min
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.
4. Applications: Limits at Infinity ★★★☆☆ ⏱ 4 min
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.
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
Why: Students forget multiplying by a negative number flips inequality signs, and do not adjust for the sign of $x$.
Why: 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.
Why: Students memorize the formula but forget it only applies when the argument of sine and the denominator both approach 0.
Why: Students rush to build bounds without starting from the known bound of the oscillating function.
Why: Students assume $x$ is positive even when the limit goes to negative infinity.