Squeeze Theorem

AP Calculus BC· AP Calculus BC CED — Limits and Continuity· 14 min read

1. What is the Squeeze Theorem?★★☆☆☆⏱ 3 min

The Squeeze Theorem (also called the Sandwich Theorem or Pinching Theorem) is a core tool for evaluating limits that cannot be solved with direct substitution, factoring, or the conjugate method. It falls in Unit 1: Limits and Continuity, which makes up 10–12% of the total AP Calculus BC exam score, and appears in both multiple-choice and free-response questions.

📘 Definition

Squeeze Theorem

\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L \implies \lim_{x \to a} f(x) = L

If for all $x$ in an open interval containing $a$ (except possibly at $a$ itself), $g (x) \leq f (x) \leq h (x)$ , and the limits of both $g (x)$ and $h (x)$ as $x \to a$ equal the same value $L$ , then the limit of $f (x)$ as $x \to a$ must also equal $L$ . The theorem adapts to one-sided limits and limits at infinity with minor domain adjustments.

Example:

Used to bound expressions involving bounded trigonometric functions to find their limits

The core intuition is simple: if $f (x)$ is trapped between two functions that both approach the same value $L$ , $f (x)$ has no choice but to also approach $L$ .

2. Applying the Squeeze Theorem to Finite Limits★★☆☆☆⏱ 4 min

The most common introductory use of the Squeeze Theorem on the AP exam is evaluating limits of bounded functions multiplied by terms that approach 0 at a finite point. We almost always leverage the universal boundedness of sine and cosine, which always have a range of $[- 1, 1]$ for any real input.

For any expression of the form $p (x) \cdot cos (g (x))$ where $lim_{x \to a} p (x) = 0$ , we can immediately write $- ∣ p (x) ∣ \leq p (x) cos (g (x)) \leq ∣ p (x) ∣$ , since $∣ cos (g (x)) ∣ \leq 1$ for any $g (x)$ . If both bounds approach 0, the entire limit is 0. This technique is also the foundation for proving the fundamental trigonometric limits $lim_{x \to 0} \frac{s i n x}{x} = 1$ and $lim_{x \to 0} \frac{1 - c o s x}{x} = 0$ , which are required for derivatives of trigonometric functions.

📐 Worked Example

Use the Squeeze Theorem to evaluate $lim_{x \to 0} x^{4} cos (\frac{3}{x ^{2}})$

1
Start with the bounded property of cosine. For all $x \neq = 0$ , the range of $cos (θ)$ is $[- 1, 1]$ , so:
$-1 \leq \cos\left(\frac{3}{x^2}\right) \leq 1$
2
Multiply all parts of the inequality by $x^{4}$ , which is non-negative for all real $x$ , so the inequality direction does not change:
$-x^4 \leq x^4 \cos\left(\frac{3}{x^2}\right) \leq x^4$
3
Evaluate the limits of the upper and lower bounding functions as $x \to 0$ :
$\lim_{x \to 0} (-x^4) = 0 \quad \text{and} \quad \lim_{x \to 0} x^4 = 0$
4
Both bounding functions approach the same limit 0, so by the Squeeze Theorem, the limit of the middle function is also 0.
$\lim_{x \to 0} x^4 \cos\left(\frac{3}{x^2}\right) = 0$

Exam tip:

Always confirm the sign of the term you multiply through the inequality. If you multiply by a negative term, you must reverse the direction of the inequality to get correct bounding functions.

3. Squeeze Theorem for Limits at Infinity★★★☆☆⏱ 3 min

For limits as $x \to \pm \infty$ , the only adjustment to the Squeeze Theorem is the domain condition: the inequality $g (x) \leq f (x) \leq h (x)$ only needs to hold for all $x$ greater than some large positive constant $M$ (for $x \to + \infty$ ) or less than some large negative constant (for $x \to - \infty$ ). The core logic remains identical: if both bounds converge to the same limit $L$ , $f (x)$ must also converge to $L$ .

The most common AP exam scenario is a bounded trigonometric numerator divided by an increasing polynomial denominator, where the whole expression is trapped between two terms that both approach 0. For sums of multiple bounded terms, we use the triangle inequality $∣ A + B ∣ \leq ∣ A ∣ + ∣ B ∣$ to find a tight upper bound for the total magnitude.

📐 Worked Example

Evaluate $lim_{x \to + \infty} \frac{3 s i n ( 2 e ^{x} ) - 5 c o s x}{x ^{2} + 4 x + 1}$ using the Squeeze Theorem.

1
Bound the numerator. For all real $x$ , $sin$ and $cos$ are bounded between $- 1$ and $1$ , so by the triangle inequality:
$|3\sin(2e^x) - 5\cos x| \leq 3|\sin(2e^x)| + 5|\cos x| \leq 3(1) + 5(1) = 8$
2
Which simplifies to the inequality:
$-8 \leq 3\sin(2e^x) - 5\cos x \leq 8$
3
The denominator $x^{2} + 4 x + 1$ is positive for all $x > 0$ , so we can divide all parts of the inequality by the denominator without changing the inequality direction:
$\frac{-8}{x^2 + 4x + 1} \leq \frac{3\sin(2e^x) - 5\cos x}{x^2 + 4x + 1} \leq \frac{8}{x^2 + 4x + 1}$
4
Evaluate the limits of the bounds as $x \to + \infty$ :
$\lim_{x \to +\infty} \frac{-8}{x^2 + 4x + 1} = 0 \quad \text{and} \quad \lim_{x \to +\infty} \frac{8}{x^2 + 4x + 1} = 0$
5
Apply the Squeeze Theorem: both bounds approach 0, so the limit of the middle function is 0.
$\lim_{x \to +\infty} \frac{3\sin(2e^x) - 5\cos x}{x^2 + 4x + 1} = 0$

Exam tip:

When bounding a sum of multiple bounded functions, always use the triangle inequality to get the maximum possible magnitude, rather than guessing a bound. This guarantees your inequality is valid for all $x$ .

4. One-Sided Limits and Continuity★★★☆☆⏱ 4 min

The Squeeze Theorem works equally well for one-sided limits ( $x \to a^{+}$ or $x \to a^{-}$ ) as it does for two-sided limits, making it a key tool for analyzing piecewise functions and confirming continuity at boundary points. For a one-sided limit, the inequality only needs to hold on the relevant side of $a$ , but the same convergence rule applies.

AP exam questions often ask to find the value of a constant that makes a piecewise function continuous at the boundary, which requires using the Squeeze Theorem to find the one-sided limits first, then matching them to find the constant. This is particularly common for piecewise functions involving absolute values.

📐 Worked Example

Let $f (x) = {\frac{( x - 2 ) ^{2} s i n ( x - 2 )}{∣ x - 2∣} C x \neq = 2 x = 2$ . Find the value of constant $C$ that makes $f (x)$ continuous at $x = 2$ , using the Squeeze Theorem.

1
For continuity at $x = 2$ , we need $lim_{x \to 2} f (x) = C$ . First evaluate the left-hand limit as $x \to 2^{-}$ : when $x < 2$ , $∣ x - 2∣ = - (x - 2)$ , so $f (x) = - (x - 2) sin (x - 2)$ . Bounding $sin (x - 2)$ gives $- 1 \leq sin (x - 2) \leq 1$ . Multiplying by $- (x - 2) = (2 - x) > 0$ gives $- (2 - x) \leq f (x) \leq (2 - x)$ . As $x \to 2^{-}$ , both bounds approach 0, so $lim_{x \to 2^{-}} f (x) = 0$ .
2
Evaluate the right-hand limit as $x \to 2^{+}$ : when $x > 2$ , $∣ x - 2∣ = x - 2$ , so $f (x) = (x - 2) sin (x - 2)$ . Bounding $sin (x - 2)$ gives the inequality:
$-(x-2) \leq (x-2)\sin(x-2) \leq (x-2)$
3
Evaluate the bounds for the right-hand limit: both $- (x - 2)$ and $(x - 2)$ approach 0 as $x \to 2^{+}$ , so $lim_{x \to 2^{+}} f (x) = 0$ .
4
The two-sided limit $lim_{x \to 2} f (x) = 0$ , so $C = 0$ to make $f$ continuous at $x = 2$ .
$C = 0$

Exam tip:

For piecewise functions with absolute values at the boundary, always split into one-sided limits first before applying the Squeeze Theorem, to ensure your bounding inequalities are correct for each side.

5. Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding of the Squeeze Theorem with this AP-style multiple choice question:

$lim_{x \to 0} x cos (\frac{2 e ^{1/ x^{2}}}{x})$ is equal to which of the following?
- A) $1$
- B) $2$
- C) $0$
- D) The limit does not exist
Reveal answer
C) $0$ —
Correct: Using the Squeeze Theorem, we get $- ∣ x ∣ \leq x cos (\dots) \leq ∣ x ∣$ , both bounds approach 0, so the limit is 0. Oscillation of the cosine term does not prevent the limit from existing when the overall expression is squeezed to 0.

6. Common Pitfalls

Wrong move:

Multiplying an inequality by a term that changes sign over the interval, and not adjusting the inequality direction.

Why:

Students often assume all powers of $x$ are positive near 0, but odd powers are negative for $x < 0$ , leading to reversed inequalities and incorrect bounds.

Correct move:

Check if your multiplier is always non-negative over the interval; if it changes sign, split into one-sided limits to handle each side separately.

Wrong move:

Only finding one bound and applying the Squeeze Theorem anyway.

Why:

Students remember a common upper bound for an expression but forget the theorem requires both upper and lower bounds to converge to the same limit.

Correct move:

Always derive both a lower bound $g (x)$ and upper bound $h (x)$ , and confirm both have the same limit before concluding the result.

Wrong move:

Using L'Hospital's Rule to prove $lim_{x \to 0} \frac{s i n x}{x} = 1$ when the question asks for a Squeeze Theorem proof.

Why:

The derivative of $sin x$ itself relies on $lim_{x \to 0} \frac{s i n x}{x} = 1$ , so this is circular reasoning that earns no credit on the AP exam.

Correct move:

Always use the method explicitly requested in the question, even if another method gives the same numerical answer.

Wrong move:

Claiming the two-sided limit exists just because the limit from the right exists.

Why:

Students forget the Squeeze Theorem for two-sided limits requires the inequality to hold on both sides of the evaluation point.

Correct move:

Always verify that your bounding inequality holds for $x < a$ and $x > a$ before concluding the two-sided limit.

Wrong move:

Using a bound that is only true for small $x$ when evaluating a limit at infinity.

Why:

Students mix up domain conditions for finite and infinite limits, leading to invalid inequalities.

Correct move:

Confirm your inequality holds for all sufficiently large (or sufficiently negative) $x$ before applying the theorem for limits at infinity.

7. Quick Reference Cheatsheet

Category	Statement/Formula	Notes
General two-sided finite limit	If $g (x) \leq f (x) \leq h (x)$ near $a$ ( $x \neq = a$ ), $lim g = lim h = L ⟹ lim f = L$	Inequality only needs to hold in an open interval around $a$
One-sided limit	Same as general, inequality holds only on one side of $a$	Use for piecewise functions and absolute value boundary problems
Limit at infinity	If $g (x) \leq f (x) \leq h (x)$ for all $x > M$ ( $x \to + \infty$ ), $lim g = lim h = L ⟹ lim f = L$	Use for bounded functions over growing denominators
Boundedness of sine/cosine	$- 1 \leq sin θ \leq 1$ ; $- 1 \leq cos θ \leq 1$	True for any real $θ$ , most common starting bound
Fundamental Trig Limit 1	$lim_{x \to 0} \frac{s i n x}{x} = 1$	Proven with Squeeze Theorem from unit circle inequalities
Fundamental Trig Limit 2	$lim_{x \to 0} \frac{1 - c o s x}{x} = 0$	Follows from first limit, can also be proven with Squeeze

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Evaluate limit of bounded function
2019 · FRQ
Find constant for continuity

What's Next

Mastering the Squeeze Theorem is a critical prerequisite for remaining topics in Unit 1: Limits and Continuity, including confirming continuity of trigonometric functions at points where they would otherwise be undefined, and evaluating indeterminate forms that cannot be solved via factoring or substitution. After Unit 1, the Squeeze Theorem is used implicitly every time you take the derivative of a sine or cosine function, since the derivatives of these core trigonometric functions rely on the fundamental trigonometric limits proven via the Squeeze Theorem. Later in the course, when studying infinite sequences and series, you will extend the Squeeze Theorem to prove convergence of bounded, monotonic sequences and evaluate limits of oscillating sequences. Bounding techniques you learn here will be useful for analysis throughout AP Calculus BC.