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Core Definition

The Squeeze Theorem (also known as the Sandwich Theorem or Pinching Theorem) states that if:

  1. g(x)≀f(x)≀h(x)g(x) \le f(x) \le h(x) for all xx in an open interval containing cc (except possibly at cc itself), and
  2. lim⁑xβ†’cg(x)=L\lim_{x \to c} g(x) = L and lim⁑xβ†’ch(x)=L\lim_{x \to c} h(x) = L,

then the trapped middle function must also converge to the exact same limit:

lim⁑xβ†’cf(x)=L\lim_{x \to c} f(x) = L

Concept Explanation

The Sandwiching Mechanism

When a function f(x)f(x) exhibits high-frequency oscillations or complex analytical forms near a boundary point cc, direct substitution and standard algebraic simplifications often fail. If two simpler bounding functions g(x)g(x) (lower bound) and h(x)h(x) (upper bound) can be constructed such that:

  1. f(x)f(x) is strictly sandwiched between g(x)g(x) and h(x)h(x), and
  2. g(x)g(x) and h(x)h(x) squeeze together to the same target value LL as x→cx \to c,

then the geometry forces f(x)f(x) into that same target LL without requiring direct algebraic resolution of f(x)f(x).

   y ^                h(x) (Upper Bound)
     |          \      /
     |           \    /  <-- Pinched together at (c, L)
   L |------------*
     |           / \     f(x) (Trapped oscillating function)
     |          /   \
     |               g(x) (Lower Bound)
     +---------------------> x
                   c

Bounding Oscillating Functions

The most common application of the Squeeze Theorem in calculus involves trigonometric functions whose arguments shoot to infinity near zero, such as sin⁑(1/x)\sin(1/x) or cos⁑(1/x)\cos(1/x).

Because the range of sine and cosine is strictly bounded by [βˆ’1,1][-1, 1]:

βˆ’1≀sin⁑(1x)≀1andβˆ’1≀cos⁑(1x)≀1(forΒ allΒ xβ‰ 0)-1 \le \sin\left(\frac{1}{x}\right) \le 1 \quad \text{and} \quad -1 \le \cos\left(\frac{1}{x}\right) \le 1 \quad (\text{for all } x \neq 0)

Multiplying this fundamental inequality by a damping envelope factor x2nx^{2n} (or ∣x∣|x|) creates the sandwich bounds (see Solving Inequalities).


Engineering Applications

  • Transient Structural Vibration Damping (Civil & Mechanical Engineering): When an earthquake or sudden mechanical shock impacts a structure, a tuned mass damper produces transient lateral sway modeled by x(t)=t2sin⁑(Ο‰/t)x(t) = t^2 \sin(\omega / t). Applying the Squeeze Theorem proves that the structural displacement converges smoothly to zero as tβ†’0+t \to 0^+ without infinite velocity shocks.
  • Optical Laser Telemetry & Diffraction (Entertainment & Computer Engineering): Laser scanning telemetry models beam envelope interference across diffraction slits. Sandwiched inequality envelopes prove beam energy thresholds remain bounded at micro-aperture centers.

Example Problems

Example 1: Damped High-Frequency Oscillation

Evaluate the limit:

lim⁑xβ†’0x2sin⁑(1x)\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right)

Solution

Step 1: Establish the trigonometric bound: For all x≠0x \neq 0, the sine function is bounded:

βˆ’1≀sin⁑(1x)≀1-1 \le \sin\left(\frac{1}{x}\right) \le 1

Step 2: Multiply by the positive damping term x2x^2: Since x2>0x^2 > 0 for all x≠0x \neq 0, multiplying preserves the inequality directions:

βˆ’x2≀x2sin⁑(1x)≀x2-x^2 \le x^2 \sin\left(\frac{1}{x}\right) \le x^2

Step 3: Evaluate the limits of the bounding functions:

lim⁑xβ†’0(βˆ’x2)=0andlim⁑xβ†’0(x2)=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: Since the lower and upper bounds both approach 00, the trapped middle function must also approach 00:

lim⁑xβ†’0x2sin⁑(1x)=0\lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0

Example 2: Inequality Bounds on an Unknown Function

Suppose a sensor signal h(t)h(t) satisfies 2t3+8≀h(t)≀t4+82t^3 + 8 \le h(t) \le t^4 + 8 for all tt near 22. Find lim⁑tβ†’2h(t)+53\lim_{t \to 2} \frac{h(t) + 5}{3}.

Solution

Step 1: Evaluate the lower and upper bound limits as t→2t \to 2:

  • Lower bound: lim⁑tβ†’2(2t3+8)=2(23)+8=16+8=24\lim_{t \to 2} (2t^3 + 8) = 2(2^3) + 8 = 16 + 8 = 24
  • Upper bound: lim⁑tβ†’2(t4+8)=24+8=16+8=24\lim_{t \to 2} (t^4 + 8) = 2^4 + 8 = 16 + 8 = 24

Step 2: Apply the Squeeze Theorem to find lim⁑tβ†’2h(t)\lim_{t \to 2} h(t): Since both bounds equal 2424, lim⁑tβ†’2h(t)=24\lim_{t \to 2} h(t) = 24.

Step 3: Evaluate the target expression using Limit Laws:

lim⁑tβ†’2h(t)+53=lim⁑tβ†’2h(t)+53=24+53=293\lim_{t \to 2} \frac{h(t) + 5}{3} = \frac{\lim_{t \to 2} h(t) + 5}{3} = \frac{24 + 5}{3} = \frac{29}{3}


Connections & References