Squeeze theorem — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The formal statement of the Squeeze Theorem (Sandwich Theorem), applications to trigonometric limits, limits at infinity, one-sided limits, and evaluating indeterminate forms involving bounded functions, aligned to the AP CED.

You should already know: Definition of one-sided and two-sided limits, basic trigonometric identities, properties of inequalities for bounded functions.

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 BC 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 (alternatively called the Sandwich Theorem or Pinching Theorem, common textbook synonyms) is a core tool for evaluating limits that cannot be found via direct substitution, factoring, or the conjugate method. According to the AP Calculus BC Course and Exam Description (CED), this topic falls within Unit 1: Limits and Continuity, which counts for 10–12% of the total AP exam weight. The Squeeze Theorem appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as a component of larger limit problems or proofs of standard trigonometric limits.

Formal definition: If for all $x$ in an open interval containing $a$ (except possibly at $a$ itself), we have $g(x)\le f(x)\le h(x)$, and ${\lim }_{x\to a}g(x)={\lim }_{x\to a}h(x)=L$, then ${\lim }_{x\to a}f(x)=L$. The theorem adapts easily to one-sided limits and limits at infinity with only minor adjustments to the domain condition (for limits as $x\to \infty$, the inequality only needs to hold for all $x>M$ for some large positive $M$, for example). The core intuition is straightforward: if $f(x)$ is trapped between two functions that both approach the same value $L$, $f(x)$ has no choice but to approach $L$ too.

2. Applying the Squeeze Theorem to Limits at Finite Points

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. The key strategy here is to leverage the universal boundedness of periodic functions like sine and cosine, which always have range $[-1,1]$ for any real input. This gives us a natural starting point to build upper and lower bounds for the function of interest. For example, for any expression of the form $p(x)\cdot \cos \left(g(x)\right)$ where ${\lim }_{x\to a}p(x)=0$, we can immediately write $-|p(x)|\le p(x)\cos (g(x))\le |p(x)|$, since $|\cos (g(x))|\le 1$ for any $g(x)$. If both $-|p(x)|$ and $|p(x)|$ approach 0 as $x\to a$, the Squeeze Theorem tells us the entire limit is 0. This technique also forms the basis for proving the two fundamental trigonometric limits ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ and ${\lim }_{x\to 0}\frac{1-\cos x}{x}=0$, which are required for all derivatives of trigonometric functions later in the course.

Worked Example

Problem: Use the Squeeze Theorem to evaluate ${\lim }_{x\to 0}{x}^4\cos \left(\frac{3}{x^2}\right)$. Step 1: Start with the bounded property of cosine. For all $x\ne 0$, the range of $\cos (\theta )$ is $[-1,1]$, so: $$-1\le \cos \left(\frac{3}{x^2}\right)\le 1$$ Step 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: $$-x^4\le x^4\cos \left(\frac{3}{x^2}\right)\le x^4$$ Step 3: Evaluate the limits of the upper and lower bounding functions as $x\to 0$: $$\underset{x\to 0}{\lim }(-x^4)=0\text{and}\underset{x\to 0}{\lim }{x}^4=0$$ Step 4: Both bounding functions approach the same limit 0, so by the Squeeze Theorem, the limit of the middle function is also 0. Result: ${\lim }_{x\to 0}{x}^4\cos \left(\frac{3}{x^2}\right)=0$

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

For limits as $x\to \pm \infty$, the Squeeze Theorem is especially useful for expressions that combine a bounded periodic function with a growing or shrinking algebraic term, which are common exam-style indeterminate forms. For limits at infinity, the only adjustment to the theorem is the domain condition: the inequality $g(x)\le f(x)\le h(x)$ 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 $|A+B|\le |A|+|B|$ to find a tight upper bound for the total magnitude of the expression.

Worked Example

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

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 with the Squeeze Theorem

The Squeeze Theorem works equally well for one-sided limits ($x\to a^{+}$ or $x\to a^{-}$) as it does for two-sided limits, which makes it a key tool for analyzing piecewise functions and confirming continuity at boundary points. For a one-sided limit as $x\to a^{+}$, the inequality $g(x)\le f(x)\le h(x)$ only needs to hold for all $x$ in the interval $(a,a+\delta )$ for some small $\delta >0$, and the same logic applies: if ${\lim }_{x\to a^{+}}g(x)={\lim }_{x\to a^{+}}h(x)=L$, then ${\lim }_{x\to a^{+}}f(x)=L$. 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, where the expression changes form on either side of the boundary.

Worked Example

Problem: Let $f(x)=\{\begin{array}{ll}\frac{(x-2{)}^2\sin (x-2)}{|x-2|}& x\ne 2\\ C& x=2\end{array}$. Find the value of constant $C$ that makes $f(x)$ continuous at $x=2$, using the Squeeze Theorem. Step 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 $-1\le \sin (x-2)\le 1$, so multiplying by $-(x-2)=(2-x)>0$ gives $-(2-x)\le f(x)\le (2-x)$. As $x\to 2^{-}$, $\lim -(2-x)=0=\lim (2-x)$, so ${\lim }_{x\to 2^{-}}f(x)=0$. Step 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 $-(x-2)\le (x-2)\sin (x-2)\le (x-2)$. Step 3: Evaluate the bounds for the right-hand limit: ${\lim }_{x\to 2^{+}}-(x-2)=0={\lim }_{x\to 2^{+}}(x-2)$, so ${\lim }_{x\to 2^{+}}f(x)=0$. Step 4: The two-sided limit ${\lim }_{x\to 2}f(x)=0$, so $C=0$ to make $f$ continuous at $x=2$.

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. Common Pitfalls (and how to avoid them)

• 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 $x<0$. 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 ${\lim }_{x\to 0}\frac{\sin x}{x}=1$ when the question asks for a Squeeze Theorem proof. Why: The derivative of $\sin x$ itself relies on ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, so this is circular reasoning that earns no credit. 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. Correct move: Confirm your inequality holds for all sufficiently large (or sufficiently negative) $x$ before applying the theorem for limits at infinity.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

${\lim }_{x\to 0}x\cos \left(\frac{2e^{1/x^2}}{x}\right)$ is equal to which of the following? A) $1$ B) $2$ C) $0$ D) The limit does not exist

Worked Solution: For any $x\ne 0$, $\cos (\theta )$ is bounded between $-1$ and $1$ for any real $\theta$, so we can write the inequality $-|x|\le x\cos \left(\frac{2e^{1/x^2}}{x}\right)\le |x|$, using the fact that $x=|x|$ for $x>0$ and $x=-|x|$ for $x<0$, which preserves the inequality direction. Next, evaluate the limits of the bounds: ${\lim }_{x\to 0}-|x|=0$ and ${\lim }_{x\to 0}|x|=0$. By the Squeeze Theorem, the limit of the middle function must be 0. Distractor D is common for students who incorrectly assume oscillation of cosine means the limit cannot exist. The correct answer is C.

Question 2 (Free Response)

Consider the function $f(x)=\frac{{\sin }^{2}x}{x}$ for $x\ne 0$. (a) Use the Squeeze Theorem to show that ${\lim }_{x\to 0}f(x)$ exists, and find its value. (b) Use your result from (a) to determine whether $f$ has a removable discontinuity at $x=0$. Justify your answer. (c) Evaluate ${\lim }_{x\to +\infty }f(x)$ using the Squeeze Theorem.

Worked Solution: (a) We know for all real $x$, $|\sin x|\le |x|$, so squaring both non-negative terms gives $0\le {\sin }^{2}x\le x^2$. For $x>0$ near 0, dividing by $x$ (positive) preserves the inequality: $0\le \frac{{\sin }^{2}x}{x}\le x$. For $x<0$ near 0, dividing by $x$ (negative) reverses the inequality: $x\le \frac{{\sin }^{2}x}{x}\le 0$. In both cases, the bounds approach 0 as $x\to 0$, so ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{+}}f(x)=0$. The two-sided limit ${\lim }_{x\to 0}f(x)=0$ exists. (b) A removable discontinuity at $x=a$ exists if ${\lim }_{x\to a}f(x)$ exists but $f(a)$ is undefined or not equal to the limit. Here, $f(0)$ is undefined, but ${\lim }_{x\to 0}f(x)=0$ exists, so $f$ has a removable discontinuity at $x=0$. (c) For all $x>0$, $0\le {\sin }^{2}x\le 1$, so dividing by $x$ (positive) gives $0\le \frac{{\sin }^{2}x}{x}\le \frac{1}{x}$. As $x\to +\infty$, ${\lim }_{x\to +\infty }0=0$ and ${\lim }_{x\to +\infty }\frac{1}{x}=0$. By the Squeeze Theorem, ${\lim }_{x\to +\infty }f(x)=0$.

Question 3 (Application / Real-World Style)

A damped mass-spring system oscillates around its equilibrium position. The displacement of the mass from equilibrium at time $t\ge 1$ seconds is given by $y(t)=\frac{\sin (20\pi t)}{\sqrt{t}}$ centimeters, where $t=0$ is the time the system is released. Use the Squeeze Theorem to find the limit of the displacement as $t\to +\infty$, and interpret your result in context.

Worked Solution: For all $t\ge 1$, $\sin (20\pi t)$ is bounded between $-1$ and $1$, so we get the inequality: $$-\frac{1}{\sqrt{t}}\le \frac{\sin (20\pi t)}{\sqrt{t}}\le \frac{1}{\sqrt{t}}$$ Evaluate the limits of the bounds: ${\lim }_{t\to +\infty }-\frac{1}{\sqrt{t}}=0$ and ${\lim }_{t\to +\infty }\frac{1}{\sqrt{t}}=0$. By the Squeeze Theorem, ${\lim }_{t\to +\infty }y(t)=0$ centimeters. This means that as time increases, the damped oscillation dies down, and the mass approaches its equilibrium position in the long run.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the Squeeze Theorem is a critical prerequisite for the next 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. Without mastering the bounding techniques required for the Squeeze Theorem, these later topics will be much harder to grasp.

Next topics to study: Limits of indeterminate forms Continuity of piecewise functions Convergence of sequences