Squeeze Theorem: Statement, How to Use It and Examples

The squeeze theorem states that if \(g(x) \le f(x) \le h(x)\) for all \(x\) near \(c\) (except possibly at \(c\)) and \(\lim_{x\to c}g(x) = \lim_{x\to c}h(x) = L\), then \(\lim_{x\to c}f(x) = L\). You use it by trapping an oscillating or tricky function between two simpler functions. Both functions’ limits at \(c\) must equal the same target value \(L\). Also known as the sandwich theorem or pinching theorem, this technique resolves limits where algebraic simplification and standard limit laws fail.

This guide explains the formal statement. It also provides a concise epsilon-delta proof and outlines a step-by-step workflow. The guide walks through six detailed examples and details the grading rubric traps that lose points on exams.

What is the squeeze theorem?

The squeeze theorem provides a way to evaluate the limit of a complicated function \(f(x)\). It does this by comparing it to two known boundary functions, \(g(x)\) and \(h(x)\). Mathematically, the theorem requires two conditions on an open interval containing \(c\) (though the equality does not need to hold at \(x = c\) itself):

1. The inequality condition: \(g(x) \le f(x) \le h(x)\) for all \(x\) near \(c\).
2. The limit condition: \(\lim_{x\to c}g(x) = \lim_{x\to c}h(x) = L\).

When both conditions are met, the limit of the inner function is guaranteed to exist and must equal \(L\):

$$\lim_{x\to c}f(x) = L$$

The theorem applies equally to left-hand and right-hand one-sided limits. It also applies to limits at infinity (\(x \to \infty\) or \(x \to -\infty\)) and infinite sequences (\(n \to \infty\)).

Why the squeeze theorem works: geometric intuition

Imagine plotting the curves of \(g(x)\) and \(h(x)\) on a graph near \(x = c\). Both curves approach the single point \((c, L)\) as \(x\) gets closer to \(c\), creating a narrowing funnel or corridor. The curve of \(f(x)\) is trapped strictly between these upper and lower walls.

Even if \(f(x)\) wiggles or oscillates infinitely fast, the walls of the corridor collapse onto the single coordinate \((c, L)\). Because \(f(x)\) cannot break through either boundary, it is forced into that exact point.

A classic analogy compares \(f(x)\) to a suspect walked between two police officers. If Officer A and Officer B both walk through the front door of a building, the suspect walking between them must go through that front door too.

The critical requirement is that the upper and lower boundaries meet at the same point. However, if the upper boundary approaches \(1\) while the lower boundary approaches \(-1\), the corridor remains wide open. This gives \(f(x)\) room to oscillate without settling on a single limit.

A formal epsilon-delta proof

Let \(\varepsilon > 0\) be any chosen tolerance. Because \(\lim_{x\to c}g(x) = L\), there exists a distance \(\delta_1 > 0\) such that whenever \(0 < |x - c| < \delta_1\), \(g(x)\) satisfies:

$$L - \varepsilon < g(x) < L + \varepsilon$$

Similarly, because \(\lim_{x\to c}h(x) = L\), there exists a distance \(\delta_2 > 0\) such that whenever \(0 < |x - c| < \delta_2\), \(h(x)\) satisfies:

$$L - \varepsilon < h(x) < L + \varepsilon$$

Choose \(\delta = \min(\delta_1, \delta_2)\). For any \(x\) satisfying \(0 < |x - c| < \delta\), both conditions hold simultaneously. This occurs alongside the trapping inequality \(g(x) \le f(x) \le h(x)\):

$$L - \varepsilon < g(x) \le f(x) \le h(x) < L + \varepsilon$$

Extracting the outer terms gives \(L - \varepsilon < f(x) < L + \varepsilon\). This means \(|f(x) - L| < \varepsilon\). Because \(\varepsilon > 0\) was arbitrary, this satisfies the formal definition of \(\lim_{x\to c}f(x) = L\).

When to use the squeeze theorem instead of other limit rules

Standard limit techniques fail in predictable scenarios where the squeeze theorem excels:

• Direct substitution fails: Expressions like \(0 \cdot \sin(1/0)\) produce undefined terms because \(\sin(1/x)\) is undefined at \(x = 0\).
• Limit laws fail: The limit product rule requires both individual limits to exist. In \(\lim_{x\to 0} x^2 \sin(1/x)\), \(\lim_{x\to 0} \sin(1/x)\) does not exist. This is because it oscillates infinitely between \(-1\) and \(1\), so you cannot split the product.
• L’Hôpital’s rule fails: Applying L’Hôpital’s rule requires the ratio of derivatives to approach a limit. Differentiating \(x^2 \sin(1/x)\) yields \(2x \sin(1/x) - \cos(1/x)\). The oscillating \(\cos(1/x)\) term still lacks a limit as \(x \to 0\).

Use the squeeze theorem whenever an expression contains an oscillating or bounded component (such as \(\sin\), \(\cos\), \((-1)^n\), or \(\arctan\)) multiplied or divided by a term that approaches zero or forces convergence.

How to use the squeeze theorem step-by-step

1. Isolate the bounded term: Find the part of the expression that oscillates but stays within fixed numerical limits. An example is \(-1 \le \sin(\text{anything}) \le 1\).
2. Build the full expression: Multiply or add across the inequality to reconstruct your original function. If you multiply by a factor that can be negative, use absolute values. Alternatively, handle positive and negative cases separately to preserve inequality signs.
3. Evaluate the boundary limits: Calculate \(\lim_{x\to c} g(x)\) and \(\lim_{x\to c} h(x)\).
4. State the equality and conclusion: Show that both outer limits equal the same value \(L\), name the theorem, and state your final limit.

A convenient variation is the absolute value squeeze theorem: if \(|f(x)| \le h(x)\) and \(\lim_{x\to c} h(x) = 0\), then \(\lim_{x\to c} f(x) = 0\). This comes from setting \(g(x) = -h(x)\). Thus, \(-h(x) \le f(x) \le h(x)\) forces \(f(x) \to 0\).

Step-by-step squeeze theorem examples

Example 1: x² sin(1/x) as x → 0

The factor \(\sin(1/x)\) oscillates wildly near \(x = 0\), but its values are always bounded between \(-1\) and \(1\):

$$-1 \le \sin\left(\frac{1}{x}\right) \le 1$$

Since \(x^2 > 0\) for all \(x \ne 0\), multiplying the entire inequality by \(x^2\) preserves the inequality directions:

$$-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2$$

Next, evaluate the limits of the outer functions as \(x \to 0\):

$$\lim_{x\to 0} (-x^2) = 0 \quad\text{and}\quad \lim_{x\to 0} x^2 = 0$$

Because both outer limits equal \(0\), the squeeze theorem dictates that \(\lim_{x\to 0} x^2 \sin(1/x) = 0\).

Example 2: x sin(1/x) as x → 0

When the multiplier is \(x\) rather than \(x^2\), \(x\) takes negative values when approaching \(0\) from the left. Multiplying an inequality by a negative number reverses the inequality signs. Avoid this complication by using absolute values:

$$\left|x \sin\left(\frac{1}{x}\right)\right| = |x| \cdot \left|\sin\left(\frac{1}{x}\right)\right| \le |x| \cdot 1 = |x|$$

Rewriting without absolute values gives:

$$-|x| \le x \sin\left(\frac{1}{x}\right) \le |x|$$

Since \(\lim_{x\to 0} (-|x|) = 0\) and \(\lim_{x\to 0} |x| = 0\), the limit is \(0\).

Example 3: cos(x)/x as x → ∞

For $x > 0$, start with the standard cosine bounds \(-1 \le \cos x \le 1\). Next, divide each part by \(x\) (which is positive as \(x \to \infty\)):

$$-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}$$

Evaluate the outer limits as \(x \to \infty\):

$$\lim_{x\to\infty} \left(-\frac{1}{x}\right) = 0 \quad\text{and}\quad \lim_{x\to\infty} \frac{1}{x} = 0$$

By the squeeze theorem, \(\lim_{x\to\infty} \frac{\cos x}{x} = 0\). This result confirms that \(y = 0\) is a horizontal asymptote for the function, a classic limit at infinity.

Example 4: Geometric squeeze proof for sin(x)/x as x → 0

The foundational trigonometric limit is:

$$\lim_{x\to 0} \frac{\sin x}{x} = 1$$

Constructing three geometric areas inside a unit circle for \(0 < x < \pi/2\) (a sector inside a triangle bounded by a larger triangle) gives the inequality \(\cos x \le \frac{\sin x}{x} \le 1\). See limit of sin x / x for the geometric diagram. Taking limits as \(x \to 0\) gives \(\lim_{x\to 0} \cos x = 1\) and \(\lim_{x\to 0} 1 = 1\). By the squeeze theorem, the limit equals \(1\). This single result establishes the derivative of \(\sin x\). It also leads directly to every trigonometric derivative, including the derivative of tan x.

Example 5: Bounded wiggle added to a growing line

Find \(\lim_{x\to\infty} \frac{2x + \sin x}{x}\).

The sine term fluctuates between \(-1\) and \(1\). Add \(2x\) across the inequality, then divide by $x > 0$:

$$\frac{2x - 1}{x} \le \frac{2x + \sin x}{x} \le \frac{2x + 1}{x}$$

Simplifying the outer fractions gives \(2 - \frac{1}{x}\) and \(2 + \frac{1}{x}\). As \(x \to \infty\), both outer expressions approach \(2\). Therefore, the limit is \(2\). A bounded oscillation becomes negligible when compared to terms growing toward infinity.

Example 6 (Exam-level): x² e^(sin(1/x)) as x → 0

Build the inequality layer by layer from the inside out:

1. Start with \(-1 \le \sin(1/x) \le 1\).
2. Apply the natural exponential function \(e^t\). Because \(e^t\) is strictly increasing, applying it preserves inequality directions: \(e^{-1} \le e^{\sin(1/x)} \le e^1\).
3. Multiply through by \(x^2 > 0\) (for \(x \ne 0\)):

$$\frac{x^2}{e} \le x^2 e^{\sin(1/x)} \le e x^2$$

Evaluating limits as \(x \to 0\) gives \(\lim_{x\to 0} \frac{x^2}{e} = 0\) and \(\lim_{x\to 0} (e x^2) = 0\). As a result, by the squeeze theorem, the limit is \(0\).

How to write squeeze theorem solutions for maximum exam credit

Calculus graders grade squeeze theorem questions against four essential steps. A full-credit solution for Example 1 requires writing:

1. State the initial bound: For all \(x \ne 0\), \(-1 \le \sin(1/x) \le 1\).
2. Justify the operation: Since \(x^2 > 0\) for all \(x \ne 0\), multiplying gives \(-x^2 \le x^2 \sin(1/x) \le x^2\).
3. Show boundary limits: \(\lim_{x\to 0} (-x^2) = 0\) and \(\lim_{x\to 0} x^2 = 0\).
4. Conclude with theorem name: Therefore, by the squeeze theorem, \(\lim_{x\to 0} x^2 \sin(1/x) = 0\).

Omitting the reason why inequality directions were preserved (stating \(x^2 > 0\)) is the single most common reason students lose points on AP and college calculus exams.

Common squeeze theorem mistakes and how to avoid them

• Bounds that do not meet: Writing \(-1 \le \sin(1/x) \le 1\) proves only that the function stays between \(-1\) and \(1\). However, it does not prove a limit exists. You must multiply by a squeezing factor that forces both bounds to the same number.
• Sign flipping errors: Multiplying an inequality by \(x\) without taking absolute values or splitting into $x > 0$ and $x < 0$ cases is invalid because negative multipliers flip inequality signs.
• Forgetting to name the theorem: Stating the outer limits meet without citing the squeeze theorem leaves your logical chain incomplete.
• Ignoring domain restrictions: For \(\lim_{x\to 0^+} \sqrt{x} \sin(1/x)\), \(\sqrt{x}\) is only defined for \(x \ge 0\). Clearly note that your squeeze holds on the one-sided interval $x > 0$.
• Splitting product limits improperly: Writing \(\lim_{x\to 0} x^2 \cdot \lim_{x\to 0} \sin(1/x)\) is mathematically illegal because the second limit does not exist.

Where the squeeze theorem appears across calculus

The core concept of trapping an unknown quantity between two convergent bounds extends far beyond limit calculations:

• Definite integrals: Lower and upper Riemann sums trap the exact area under a curve. As subinterval widths approach zero, both sums are squeezed to the definite integral value.
• Fundamental Theorem of Calculus: Proving the fundamental theorem of calculus involves squeezing the average value of a continuous function between its minimum and maximum values on an infinitesimal interval.
• Sequences and series: Testing sequence convergence often uses sequence bounds, such as proving \(\lim_{n\to\infty} \frac{\cos n}{n} = 0\) via \(-\frac{1}{n} \le \frac{\cos n}{n} \le \frac{1}{n}\).

See the squeeze theorem on a graph

In this interactive plot, \(x \sin(1/x)\) oscillates rapidly near \(x = 0\), trapped perfectly between the boundary lines \(y = x\) and \(y = -x\):

To double-check work on complex functions, use the limit calculator, or run one-sided calculations with the one-sided limit calculator when bounds depend on direction.

Practice problems with full solutions

1. Evaluate \(\lim_{x\to 0} x^4 \cos\left(\frac{2}{x}\right)\)
2. Next, Evaluate \(\lim_{x\to\infty} \frac{\sin(x^2)}{x}\)
3. Evaluate \(\lim_{x\to 0^+} \sqrt{x} \sin\left(\frac{1}{x}\right)\)
4. Also, Evaluate \(\lim_{x\to\infty} \frac{3x^2 + \cos x}{x^2}\)
5. Evaluate \(\lim_{x\to\infty} \frac{\sin x}{e^x}\)
6. Then, Evaluate \(\lim_{x\to 0} x^3 \sin\left(\frac{1}{x}\right)\)

Solutions:

1. 0: Bound as \(-x^4 \le x^4 \cos(2/x) \le x^4\). Both bounds approach \(0\) as \(x \to 0\).
2. 0: Bound for $x > 0$ as \(-\frac{1}{x} \le \frac{\sin(x^2)}{x} \le \frac{1}{x}\). Both bounds approach \(0\) as \(x \to \infty\).
3. 0: Bound for $x > 0$ as \(-\sqrt{x} \le \sqrt{x} \sin(1/x) \le \sqrt{x}\). Both bounds approach \(0\) as \(x \to 0^+\).
4. 3: Split fraction to get \(3 + \frac{\cos x}{x^2}\). As a result, bound as \(3 - \frac{1}{x^2} \le \frac{3x^2 + \cos x}{x^2} \le 3 + \frac{1}{x^2}\). Both bounds approach \(3\) as \(x \to \infty\).
5. 0: Bound as \(-e^{-x} \le \frac{\sin x}{e^x} \le e^{-x}\). Both bounds approach \(0\) as \(x \to \infty\).
6. 0: Use absolute values: \(-|x|^3 \le x^3 \sin(1/x) \le |x|^3\). Both bounds approach \(0\) as \(x \to 0\).

Frequently asked questions

Is the squeeze theorem the same as the sandwich theorem?

Yes. Squeeze theorem, sandwich theorem, and pinching theorem are synonymous terms for the exact same mathematical theorem.

Does f(x) have to be trapped between g(x) and h(x) at x = c?

No. Limits describe function behavior near \(c\), not at \(c\). The inequality \(g(x) \le f(x) \le h(x)\) must hold on an open interval around \(c\), but \(f(c)\) does not even need to be defined.

Can the squeeze theorem prove that a limit does not exist?

No. The squeeze theorem only proves existence and finds the value when outer limits match. If your boundary functions approach different values, the test is inconclusive and provides no information.

Does the squeeze theorem work for infinite sequences?

Yes. If \(a_n \le b_n \le c_n\) for all \(n\) beyond some integer \(N\), and \(\lim_{n\to\infty} a_n = \lim_{n\to\infty} c_n = L\), then \(\lim_{n\to\infty} b_n = L\).

How do I choose the upper and lower bounding functions?

Isolate the component with known minimum and maximum values (like \(-1 \le \sin\theta \le 1\)), then perform identical operations across the inequality to reconstruct the full target expression.

Further reading

• Computing Limits (Paul’s Online Math Notes) — the squeeze theorem statement with a graph and a worked \(x^2 \cos x\) example, alongside other limit techniques.
• Squeeze theorem (Wikipedia) — the formal statement, a proof, and versions for sequences and several variables.