The Squeeze Theorem

Master the art of bounding a function between two simpler ones to determine limits that resist direct computation.

AP Calculus BC

What Is the Squeeze Theorem?

Sometimes a function oscillates or behaves erratically near a point, making it impossible to evaluate the limit by direct substitution, factoring, or algebraic manipulation. The Squeeze Theorem offers an elegant workaround: if you can trap the troublesome function between two bounding functions that converge to the same limit, the squeezed function must converge there too.

The Bounding Idea

Upper Bound

A function \( h(x) \) that is always greater than or equal to \( f(x) \) near the point of interest.

Target

The function \( f(x) \) whose limit we want but cannot evaluate directly.

Lower Bound

A function \( g(x) \) that is always less than or equal to \( f(x) \) near the point of interest.

Visual Intuition

Formal Statement

The theorem also applies to one-sided limits (\( x \to a^+ \) or \( x \to a^- \)) and limits at infinity (\( x \to \infty \)). The inequality only needs to hold near the point, not everywhere.

Three Conditions to Verify

1

Establish the inequality. Show that \( g(x) \leq f(x) \leq h(x) \) near \( x = a \). Usually comes from a known bound like \( -1 \leq \sin(\theta) \leq 1 \).

2

Compute the outer limits. Evaluate \( \lim_{x \to a} g(x) \) and \( \lim_{x \to a} h(x) \) separately.

3

Confirm they agree. If both limits equal the same value \( L \), conclude \( \lim_{x \to a} f(x) = L \) by the Squeeze Theorem.

Squeeze Theorem vs. Direct Substitution

	Squeeze Theorem	Direct Substitution
When to use	Function oscillates or is undefined at \( a \)	Function is continuous at \( a \)
Requires	Two bounding functions with matching limits	Simply plug in \( x = a \)
Typical signals	\( \sin(1/x) \), \( \cos(1/x) \), bounded oscillation	Polynomial, rational (no hole), exponential
Difficulty	Finding the right bounds	Minimal

How to Apply Step by Step

1

Identify the oscillating factor. Look for bounded terms like \( \sin(\cdot) \), \( \cos(\cdot) \), or \( (-1)^n \).

2

Write the known bound. For example, \( -1 \leq \sin\!\left(\frac{1}{x}\right) \leq 1 \).

3

Multiply through by the damping factor. If the full expression is \( x^2 \sin(1/x) \), multiply the bound by \( x^2 \) to get \( -x^2 \leq x^2 \sin(1/x) \leq x^2 \).

4

Evaluate the outer limits. Both \( \lim_{x\to 0}(-x^2) = 0 \) and \( \lim_{x\to 0} x^2 = 0 \).

5

Conclude. By the Squeeze Theorem, the limit equals \( 0 \).

Example 1: \( \lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) \)

Example 2: \( \lim_{x \to 0} x\cos\!\left(\frac{1}{x}\right) \)

1

Bound the oscillating factor: \( -1 \leq \cos(1/x) \leq 1 \).

2

Multiply by \( |x| \): Since \( x \) changes sign, use absolute values: \( -|x| \leq x\cos(1/x) \leq |x| \).

3

Outer limits: \( \lim_{x\to 0}(-|x|) = 0 \) and \( \lim_{x\to 0}|x| = 0 \).

4

Conclude: By the Squeeze Theorem, \( \lim_{x\to 0} x\cos(1/x) = 0 \).

Example 3: \( \lim_{x \to \infty} \frac{\sin x}{x} \)

When Bounds Don't Match

The Squeeze Theorem only works when the upper and lower bounds converge to the same limit. If \( \lim g(x) \neq \lim h(x) \), the theorem tells us nothing about \( \lim f(x) \): the limit may exist at any value between them, or may not exist at all.

Classic Bound Toolkit

\( \sin \) and \( \cos \)

\( -1 \leq \sin\theta \leq 1 \) and \( -1 \leq \cos\theta \leq 1 \) for all \( \theta \).

\( \sin^2 \) and \( \cos^2 \)

\( 0 \leq \sin^2\theta \leq 1 \) and \( 0 \leq \cos^2\theta \leq 1 \).

Floor / Fractional

\( x - 1 < \lfloor x \rfloor \leq x \), so \( \frac{\lfloor x \rfloor}{x} \) can be squeezed.

The Role of the Damping Factor

The Squeeze Theorem typically involves two pieces: a bounded oscillating factor (like \( \sin(1/x) \)) and a damping factor that drives the product to zero (like \( x^2 \)). The damping factor is what makes the bounds converge.

Oscillating Factor

Stays bounded but never settles: \( \sin(1/x) \), \( \cos(e^x) \), \( (-1)^n \).

Damping Factor

Goes to zero and pulls the product with it: \( x^2 \), \( 1/x \), \( e^{-x} \).

Common Mistakes

✕ Bounds Don't Match

Applying the theorem when \( \lim g(x) \neq \lim h(x) \). The limits of both bounds must be equal.

✕ Sign Errors

When multiplying an inequality by a negative quantity, forgetting to flip the inequality signs.

✕ Wrong Interval

The inequality \( g(x) \leq f(x) \leq h(x) \) must hold near the limit point, not just at isolated values.

Summary of Key Limits

Limit Expression	Lower Bound	Upper Bound
\( x^2\sin(1/x) \) as \( x\to 0 \)	\( -x^2 \)	\( x^2 \)
\( x\cos(1/x) \) as \( x\to 0 \)	\( -\|x\| \)	\( \|x\| \)
\( \frac{\sin x}{x} \) as \( x\to\infty \)	\( -\frac{1}{x} \)	\( \frac{1}{x} \)
\( e^{-x}\cos(x) \) as \( x\to\infty \)	\( -e^{-x} \)	\( e^{-x} \)

Question 1 of 5

What is \( \displaystyle\lim_{x \to 0}\, x^2 \cos\!\left(\frac{1}{x}\right) \)?

Since \( -1 \leq \cos(1/x) \leq 1 \), we get \( -x^2 \leq x^2\cos(1/x) \leq x^2 \). Both \( \pm x^2 \to 0 \) as \( x \to 0 \), so by the Squeeze Theorem the limit is \( 0 \).

Question 2 of 5

Which condition is not required for the Squeeze Theorem?

The Squeeze Theorem does not require \( f(a) \) to be defined. The inequality and the matching limits of the bounds need only hold near \( a \), not necessarily at \( a \).

Question 3 of 5

To use the Squeeze Theorem on \( \displaystyle\lim_{x\to\infty} \frac{\cos(x^2)}{x} \), which pair of bounds is correct?

Since \( -1 \leq \cos(x^2) \leq 1 \), dividing by \( x > 0 \) gives \( -\frac{1}{x} \leq \frac{\cos(x^2)}{x} \leq \frac{1}{x} \). Option (b) is wrong because cosine can be negative. Option (d) is too tight: the bound is \( 1/x \), not \( 1/x^2 \).

Question 4 of 5

If \( 3x - 1 \leq f(x) \leq x^2 + 3 \) for all \( x \) near \( 2 \), what is \( \displaystyle\lim_{x \to 2} f(x) \)?

At \( x = 2 \): \( g(2) = 3(2)-1 = 5 \) and \( h(2) = 2^2+3 = 7 \). Since \( 5 \neq 7 \), the two bounds do not converge to the same limit. The Squeeze Theorem cannot be applied, so we cannot determine the limit from this information alone.

Question 5 of 5

What is \( \displaystyle\lim_{x\to 0^+} \sqrt{x}\,\sin\!\left(\frac{1}{x}\right) \)?

For \( x > 0 \), \( \sqrt{x} > 0 \) and \( -1 \leq \sin(1/x) \leq 1 \), so \( -\sqrt{x} \leq \sqrt{x}\sin(1/x) \leq \sqrt{x} \). Since both \( \pm\sqrt{x} \to 0 \) as \( x \to 0^+ \), the Squeeze Theorem gives the limit as \( 0 \).