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Squeeze Theorem

The Squeeze Theorem says that if a function f(x)f(x) is trapped between two other functions that both approach the same limit LL, then f(x)f(x) must also approach LL. It's a way to find limits that are difficult or impossible to compute directly.

Let g(x)g(x), f(x)f(x), and h(x)h(x) be functions defined on an open interval containing cc (except possibly at cc itself). If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) for all xx in that interval and lim⁡x→cg(x)=lim⁡x→ch(x)=L\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, then lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L. The theorem works because f(x)f(x) is "squeezed" between two functions that converge to the same value, leaving it no room to go anywhere else.

Key Formula

If g(x)≤f(x)≤h(x) and lim⁡x→cg(x)=lim⁡x→ch(x)=L, then lim⁡x→cf(x)=L.\text{If } g(x) \le f(x) \le h(x) \text{ and } \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, \text{ then } \lim_{x \to c} f(x) = L.
Where:
  • f(x)f(x) = the function whose limit you want to find
  • g(x)g(x) = a lower bounding function
  • h(x)h(x) = an upper bounding function
  • LL = the common limit of the bounding functions
  • cc = the value that x approaches

Worked Example

Problem: Find lim⁡x→0x2sin⁡ ⁣(1x)\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right).
Step 1: Recognize the difficulty. As x→0x \to 0, sin⁡(1/x)\sin(1/x) oscillates wildly between −1-1 and 11, so you can't evaluate the limit by direct substitution. This is a perfect candidate for the Squeeze Theorem.
Step 2: Set up the inequality. Since −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for all x≠0x \ne 0, multiply every part by x2x^2 (which is always non-negative, so the inequality direction is preserved).
−x2≤x2sin⁡ ⁣(1x)≤x2-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2
Step 3: Find the limits of the outer functions as x→0x \to 0.
lim⁡x→0(−x2)=0andlim⁡x→0x2=0\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0
Step 4: Apply the Squeeze Theorem. Both bounding functions approach 00, and x2sin⁡(1/x)x^2 \sin(1/x) is trapped between them, so:
lim⁡x→0x2sin⁡ ⁣(1x)=0\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0
Answer: The limit is 00.

Visualization

Why It Matters

Many important limits in calculus cannot be computed by substitution or algebraic simplification — the classic example being lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1, which is proved using the Squeeze Theorem with geometric bounds from a unit circle. In physics and engineering, the theorem helps analyze oscillating systems where a quantity is bounded by two simpler expressions that converge to the same value.

Common Mistakes

Mistake: Trying to apply the theorem when the two outer limits are not equal.
Correction: The Squeeze Theorem only works when lim⁡g(x)\lim g(x) and lim⁡h(x)\lim h(x) are the same value LL. If they differ, you cannot conclude anything about lim⁡f(x)\lim f(x).
Mistake: Flipping the inequality when multiplying by a negative or variable expression.
Correction: When you multiply an inequality by a factor, check its sign. Multiplying by a negative quantity reverses the inequality. In the worked example, x2≥0x^2 \ge 0 for all xx, so the direction stays the same — always verify this.

Related Terms

  • Pinching Theorem — Another name for the Squeeze Theorem
  • Limit — The fundamental concept the theorem evaluates