1.8 Determining Limits Using the Squeeze Theorem

The Squeeze Theorem lets you find a limit by bounding a tricky function between two simpler functions that approach the same value. If $f(x)\leq g(x)\leq h(x)$ near a point and both $f$ and $h$ approach the same limit $L$, then $g$ must also approach $L$. For AP Calculus, state the shared outer limit before concluding the middle limit.

Why This Matters for the AP Calculus Exam

The Squeeze Theorem shows up when a limit cannot be found by direct substitution, factoring, or other algebraic moves. It is especially useful for oscillating functions like $\sin\left(\frac{1}{x}\right)$ and for the special trig limits that come back later in derivatives.

On the AP Calculus exam, you can expect to use this idea in both multiple-choice questions and free-response questions. Some problems give you bounding inequalities and ask you to draw a conclusion about a "squeezed" function. To support a stronger score on free-response work, you need to state that the bounding functions share the same limit before concluding the middle function shares it too. Clear notation and showing that the conditions are met are important for clear exam work.

more resources to help you study

practice multiple choiceFRQ practice & scoringcheatsheetsscore calculatorkey terms

Key Takeaways

• The Squeeze Theorem applies when $f(x) \leq g(x) \leq h(x)$ near a point and $\lim f(x) = \lim h(x) = L$, which forces $\lim g(x) = L$.
• Use it when you cannot evaluate a limit directly, especially for bounded oscillating functions.
• Two classic results come from this theorem: $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{1 - \cos x}{x} = 0$.
• A common bounding move uses $-1 \leq \cos\theta \leq 1$ or $-1 \leq \sin\theta \leq 1$ to trap a product like $x\cos\left(\frac{1}{x}\right)$.
• Always confirm the outer functions actually approach the same value before you state the limit of the inner function.
• State your reasoning explicitly: show the inequality, show both outer limits, then conclude.

How the Squeeze Theorem Works

The Squeeze Theorem states that if

$f(x) \leq g(x) \leq h(x)$

near $x = a$ and

$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L,$

then

$\lim_{x \to a} g(x) = L.$

The idea is visual: if $g(x)$ is always sandwiched between $f(x)$ and $h(x)$, and the top and bottom functions both squeeze toward the same value $L$, then $g(x)$ has nowhere else to go. It gets pinned to $L$ too.

Background You Need

To use the Squeeze Theorem smoothly, get comfortable with:

• Limits: how a function behaves near a specific input value, including one-sided behavior.
• Bounded functions: knowing that sine and cosine always stay between $-1$ and $1$, which gives you ready-made inequalities to build with.

How to Use This on the AP Calculus Exam

Problem Solving

A reliable order of steps:

1. Identify the function whose limit you need.
2. Find the squeeze functions. Build a lower bound $f(x)$ and upper bound $h(x)$ using known inequalities, often $-1 \leq \cos\theta \leq 1$ or $-1 \leq \sin\theta \leq 1$.
3. Check the outer limits. Confirm that $f$ and $h$ approach the same value as $x$ approaches the target.
4. Conclude. State that the squeezed function shares that limit.

Free Response

Some free-response problems hand you the inequality and ask you to justify a conclusion. Here is the type of reasoning they want.

Functions $g$ and $h$ are twice-differentiable with $g(2) = h(2) = 4$, and $g(x) < h(x)$ for $1 < x < 3$. Let $k$ satisfy $g(x) \leq k(x) \leq h(x)$ for $1 < x < 3$. Is $k$ continuous at $x = 2$?

Because $g$ and $h$ are twice-differentiable, they are continuous, so $\lim_{x \to 2} g(x) = 4$ and $\lim_{x \to 2} h(x) = 4$. Since $g(x) \leq k(x) \leq h(x)$, the Squeeze Theorem gives $\lim_{x \to 2} k(x) = 4$. The inequality also forces $k(2) = 4$. Because $\lim_{x \to 2} k(x) = k(2) = 4$, $k$ is continuous at $x = 2$. For more on the continuity conditions, see Confirming Continuity Over an Interval.

This problem comes from the 2019 AP Calculus AB exam released by College Board.

Worked Example: Computing a Limit

Find $\lim_{x \to 0} x\cos\left(\frac{1}{x}\right)$ using the Squeeze Theorem.

Start with the bounded factor:

$-1 \leq \cos\left(\frac{1}{x}\right) \leq 1$

Multiply through by $x$ (using absolute-value bounds to keep the inequality safe near $0$) to get bounding functions $f(x) = -|x|$ and $h(x) = |x|$, so that $x\cos\left(\frac{1}{x}\right)$ stays between them. Then check the outer limits:

$\lim_{x \to 0} (-|x|) = 0 \qquad \lim_{x \to 0} |x| = 0$

Since both outer functions approach $0$,

$\lim_{x \to 0} x\cos\left(\frac{1}{x}\right) = 0.$

Common Trap

When you multiply an inequality by $x$, the direction of the inequality flips for negative $x$. Using $-|x|$ and $|x|$ as your bounds keeps the trapping valid on both sides of $0$, which matters for a two-sided limit.

Common Misconceptions

• The outer functions do not need to equal each other everywhere, only share the same limit. $f$ and $h$ can be different functions; what matters is that both approach the same $L$ at the target point.
• You cannot skip checking both outer limits. If $f$ and $h$ approach different values, the theorem tells you nothing about the middle function.
• The squeeze must hold near the point, not just at it. You need the inequality $f(x) \leq g(x) \leq h(x)$ on an interval around the point (the point itself can be excluded), not at a single value.
• A bounded function alone does not make a limit zero. $\cos\left(\frac{1}{x}\right)$ stays bounded but has no limit at $0$. It is the product with a factor going to $0$, like $x$, that gets squeezed to $0$.
• Plain substitution will not work here. These limits resist direct substitution, which is exactly why the Squeeze Theorem is the right tool.

Related AP Calculus Guides

• 1.2 Defining Limits and Using Limit Notation
• 1.1 Introducing Calculus: Can Change Occur at An Instant?
• 1.6 Determining Limits Using Algebraic Manipulation
• 1.3 Estimating Limit Values from Graphs
• 1.5 Determining Limits Using Algebraic Properties of Limits
• Unit 1 Overview: Limits and Continuity