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1.8 Determining Limits Using the Squeeze Theorem

Keywords

Squeeze Theorem (Sandwich Theorem) 夹逼定理 • Geometric Proof 几何证明

Unit Circle 单位圆 • Indeterminate Trig Limits 三角不定式 • Conjugate Multiplication 共轭相乘

The Squeeze Theorem

The Squeeze Theorem allows us to find the limit of a "difficult" function by trapping it between two "easier" functions that share the same limit at a specific point.

核心要点：The Squeeze Theorem

Suppose \(g(x) \leq f(x) \leq h(x)\) for all \(x\) in an open interval containing \(c\), except possibly at \(c\) itself.

If:

\[ \lim_{x \to c} g(x) = L \quad \text{and} \quad \lim_{x \to c} h(x) = L \]

Then:

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

Visualizing the Squeeze

In the graph below, notice how \(f(x) = x^2 \sin(1/x)\) (in blue) is "squeezed" by \(h(x) = x^2\) and \(g(x) = -x^2\) as \(x \to 0\).

Because \(\lim\limits_{x \to 0} (-x^2) = 0\) and \(\lim\limits_{x \to 0} (x^2) = 0\), the function in the middle must also have a limit of \(0\).

Proof 1: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\)

This limit is the "Holy Grail" of trigonometric limits. We cannot use direct substitution (gives \(0/0\)) or simple algebra. We must use a geometric squeeze.

1. The Setup

Consider a unit circle (radius \(r=1\)) and an angle \(x\) in the first quadrant (\(0 < x < \pi/2\)). We compare the areas of three distinct regions:

  1. Inner Triangle: A triangle with base \(1\) and height \(\sin x\). Area \(= \frac{1}{2}(1)(\sin x) \cos x = \frac{1}{2} \sin x \cos x\).

  2. Circular Sector: A slice of the circle with angle \(x\). Area \(= \frac{1}{2}r^2\theta = \frac{1}{2}(1)^2 x = \frac{1}{2}x\).

  3. Outer Triangle: A larger triangle with base \(1\) and height \(\tan x\). Area \(= \frac{1}{2}(1)(\tan x) = \frac{1}{2} \tan x\).

2. The Inequality

Visually, the Inner Triangle \(\subset\) Circular Sector \(\subset\) Outer Triangle. Thus:

\[ \frac{1}{2} \sin x \cos x < \frac{1}{2} x < \frac{1}{2} \tan x \]

Multiply everything by \(2\) and divide by \(\sin x\) (which is positive for \(x \to 0^+\)):

\[ \cos x < \frac{x}{\sin x} < \frac{1}{\cos x} \]

3. The Squeeze

Taking the reciprocal (which flips the inequality signs):

\[ \frac{1}{\cos x} > \frac{\sin x}{x} > \cos x \]

As \(x \to 0\):

\[ \lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{1}{\cos x} = 1 \]

By the Squeeze Theorem, the middle term must also approach 1.

Proof 2: \(\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\)

Now that we know \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), we can use algebra and the conjugate to prove this second special limit.

  1. Multiply by the Conjugate: Multiply the numerator and denominator by \((1 + \cos x)\).

    \[ \frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} = \frac{1 - \cos^2 x}{x(1 + \cos x)} \]

  2. Apply Pythagorean Identity: Recall that \(1 - \cos^2 x = \sin^2 x\).

    \[ \frac{\sin^2 x}{x(1 + \cos x)} = \left( \frac{\sin x}{x} \right) \cdot \left( \frac{\sin x}{1 + \cos x} \right) \]

  3. Evaluate the Limits: Apply the limit as \(x \to 0\) to both parts:

    \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
    \[ \lim_{x \to 0} \frac{\sin x}{1 + \cos x} = \frac{0}{1 + 1} = 0 \]

  4. Result: \(1 \cdot 0 = 0\).

Summary of Special Limits

Important Limits to Memorize

You will use these repeatedly to solve complex trigonometric limits:

  1. \[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
  2. \[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

Quiz