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Unit 1.8 – Determining Limits Using the Squeeze Theorem

AP® Calculus AB & BC | Formula Reference Sheet

Core Concept: The Squeeze Theorem (also called the Sandwich Theorem or Pinching Theorem) is a powerful tool for finding limits of functions that are "trapped" between two simpler functions. If you can't evaluate a limit directly, squeeze it!

🎯 The Squeeze Theorem Statement

THE SQUEEZE THEOREM (SANDWICH THEOREM)

If three functions \(g(x)\), \(f(x)\), and \(h(x)\) satisfy:

\[ g(x) \leq f(x) \leq h(x) \]

for all \(x\) near \(a\) (except possibly at \(a\) itself), and if:

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

Then:

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

📝 Intuition: Think of \(f(x)\) as being "sandwiched" between \(g(x)\) (bottom slice of bread) and \(h(x)\) (top slice of bread). If both slices meet at the same point \(L\), the filling \(f(x)\) has nowhere else to go—it must also approach \(L\)!

👁️ Visual Understanding

Imagine three functions graphed:

  • \(g(x)\) (lower bound): Blue curve on the bottom
  • \(f(x)\) (middle function): Red curve in the middle
  • \(h(x)\) (upper bound): Green curve on top

As \(x \to a\), if both blue and green curves approach the same height \(L\), the red curve (trapped between them) must also approach \(L\).

🔍 When to Use the Squeeze Theorem

Use the Squeeze Theorem when:

  • Oscillating functions: Functions like \(\sin(1/x)\) or \(\cos(1/x)\) that wiggle infinitely near a point
  • Bounded functions times vanishing terms: Like \(x \sin(1/x)\) or \(x^2 \cos(1/x)\)
  • Can't simplify algebraically: When factoring, conjugates, and other methods fail
  • Known inequalities exist: When you can establish bounds using trig identities or algebraic inequalities

📝 Step-by-Step Process

How to Apply the Squeeze Theorem:

  1. Identify the "squeezed" function: This is \(f(x)\)—the function whose limit you need
  2. Find bounding functions: Find \(g(x)\) and \(h(x)\) such that \(g(x) \leq f(x) \leq h(x)\)
  3. Verify the inequality: Prove or justify that the inequality holds for all \(x\) near \(a\)
  4. Evaluate the limits of bounds: Calculate \(\lim_{x \to a} g(x)\) and \(\lim_{x \to a} h(x)\)
  5. Check if limits are equal: If both equal \(L\), proceed
  6. Apply the theorem: Conclude that \(\lim_{x \to a} f(x) = L\)

📐 Essential Inequalities to Know

Fundamental Trig Inequalities

These are CRUCIAL for AP® Calculus—memorize them!

\[ -1 \leq \sin(x) \leq 1 \quad \text{for all } x \]
\[ -1 \leq \cos(x) \leq 1 \quad \text{for all } x \]
\[ \cos(x) \leq \frac{\sin(x)}{x} \leq 1 \quad \text{for } 0 < |x| < \frac{\pi}{2} \]

💡 Pro Tip: When you see \(\sin(\text{anything})\) or \(\cos(\text{anything})\), immediately think: "This is bounded between -1 and 1!" This is the key to most squeeze theorem problems.

⭐ The Two Most Important Limits

Limit 1: sin(x)/x as x → 0

The Famous Sine Limit
\[ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \]

Proof Using Squeeze Theorem:

  1. Start with the inequality: For \(0 < x < \frac{\pi}{2}\):
    \[ \cos(x) \leq \frac{\sin(x)}{x} \leq 1 \]
  2. This comes from geometry: Using a unit circle and comparing areas of triangles and sectors
  3. Evaluate the bounds:
    • \(\lim_{x \to 0} \cos(x) = 1\)
    • \(\lim_{x \to 0} 1 = 1\)
  4. By Squeeze Theorem: \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) ✓

Limit 2: (1 - cos(x))/x as x → 0

The Cosine Limit
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0 \]

Proof Using Squeeze Theorem:

  1. Use identity: \(1 - \cos(x) = 2\sin^2(x/2)\)
  2. Rewrite:
    \[ \frac{1-\cos(x)}{x} = \frac{2\sin^2(x/2)}{x} = \frac{\sin(x/2)}{x/2} \cdot \sin(x/2) \]
  3. As \(x \to 0\):
    • \(\frac{\sin(x/2)}{x/2} \to 1\) (by Limit 1)
    • \(\sin(x/2) \to 0\)
  4. Result: \(1 \cdot 0 = 0\) ✓

⚠️ MEMORIZE THESE TWO! These limits appear constantly in calculus. You should know them by heart:

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

📚 Worked Examples

Example 1: Classic Squeeze Problem

Find: \(\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)\)

Solution:

  1. Note: Direct substitution gives \(0 \cdot \sin(\infty)\) which is indeterminate
  2. Key inequality: \(-1 \leq \sin\left(\frac{1}{x}\right) \leq 1\)
  3. Multiply by \(|x|\): Since \(x \to 0\), consider both positive and negative:
    • For \(x > 0\): \(-x \leq x\sin\left(\frac{1}{x}\right) \leq x\)
    • For \(x < 0\): \(x \leq x\sin\left(\frac{1}{x}\right) \leq -x\)
  4. General form: \(-|x| \leq x\sin\left(\frac{1}{x}\right) \leq |x|\)
  5. Evaluate bounds:
    • \(\lim_{x \to 0} (-|x|) = 0\)
    • \(\lim_{x \to 0} |x| = 0\)
  6. By Squeeze Theorem: \(\lim_{x \to 0} x\sin\left(\frac{1}{x}\right) = 0\)

Answer: 0

Example 2: Polynomial with Sine

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

Solution:

  1. Key inequality: \(-1 \leq \cos\left(\frac{1}{x}\right) \leq 1\)
  2. Multiply by \(x^2\): \(-x^2 \leq x^2\cos\left(\frac{1}{x}\right) \leq x^2\)
  3. Note: \(x^2 \geq 0\), so inequality signs don't flip
  4. Evaluate bounds:
    • \(\lim_{x \to 0} (-x^2) = 0\)
    • \(\lim_{x \to 0} x^2 = 0\)
  5. By Squeeze Theorem: \(\lim_{x \to 0} x^2\cos\left(\frac{1}{x}\right) = 0\)

Answer: 0

Example 3: Given Inequality

Find: \(\lim_{x \to 0} f(x)\) if \(4x - 9 \leq f(x) \leq x^2 - 4x + 7\) for \(x > 0\)

Solution:

  1. Already have bounds: \(g(x) = 4x - 9\) and \(h(x) = x^2 - 4x + 7\)
  2. Evaluate lower bound: \(\lim_{x \to 0} (4x - 9) = -9\)
  3. Evaluate upper bound: \(\lim_{x \to 0} (x^2 - 4x + 7) = 7\)
  4. Problem! Limits are NOT equal (\(-9 \neq 7\))
  5. Conclusion: Cannot use Squeeze Theorem with these bounds!

Important: Both bounding limits MUST be equal for the theorem to work!

Example 4: Using sin(x)/x Formula

Find: \(\lim_{x \to 0} \frac{\sin(3x)}{5x}\)

Solution:

  1. Rewrite to match pattern:
    \[ \frac{\sin(3x)}{5x} = \frac{3}{5} \cdot \frac{\sin(3x)}{3x} \]
  2. Apply famous limit: As \(x \to 0\), \(3x \to 0\), so:
    \[ \lim_{x \to 0} \frac{\sin(3x)}{3x} = 1 \]
  3. Result: \(\frac{3}{5} \cdot 1 = \frac{3}{5}\)

Answer: \(\frac{3}{5}\)

Example 5: Limit at Infinity

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

Solution:

  1. Key inequality: \(-1 \leq \sin(x) \leq 1\)
  2. Divide by \(x\) (positive for large \(x\)):
    \[ -\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x} \]
  3. Evaluate bounds:
    • \(\lim_{x \to \infty} \left(-\frac{1}{x}\right) = 0\)
    • \(\lim_{x \to \infty} \frac{1}{x} = 0\)
  4. By Squeeze Theorem: \(\lim_{x \to \infty} \frac{\sin(x)}{x} = 0\)

Answer: 0

🎯 Common Squeeze Theorem Patterns

Pattern Key Inequality Typical Limit
\(x\sin(1/x)\) \(-|x| \leq x\sin(1/x) \leq |x|\) \(\lim_{x \to 0} = 0\)
\(x^n\sin(1/x)\) \(-|x|^n \leq x^n\sin(1/x) \leq |x|^n\) \(\lim_{x \to 0} = 0\)
\(x^2\cos(1/x)\) \(-x^2 \leq x^2\cos(1/x) \leq x^2\) \(\lim_{x \to 0} = 0\)
\(\sin(x)/x\) \(\cos(x) \leq \frac{\sin(x)}{x} \leq 1\) \(\lim_{x \to 0} = 1\)
\(\sin(x)/x\) at \(\infty\) \(-\frac{1}{x} \leq \frac{\sin(x)}{x} \leq \frac{1}{x}\) \(\lim_{x \to \infty} = 0\)

💡 Tips, Tricks & Strategies

✅ Essential Tips

  • Recognize bounded functions: \(\sin\) and \(\cos\) are ALWAYS between -1 and 1
  • Look for "vanishing" terms: If bounded function is multiplied by something → 0, use squeeze!
  • Be careful with signs: When multiplying inequalities, watch for negative numbers
  • Both bounds must converge to SAME value: If they don't, squeeze doesn't work
  • Justify your inequality: On AP® exams, you must explain WHY the inequality holds

🎯 Strategy: Finding Bounding Functions

How to find \(g(x)\) and \(h(x)\):

  1. Identify the bounded part: Usually \(\sin\) or \(\cos\)
  2. Use standard inequality: \(-1 \leq \sin(\text{anything}) \leq 1\)
  3. Multiply by the rest: Whatever multiplies the bounded function
  4. Simplify bounds: Make sure \(g(x)\) and \(h(x)\) are simple enough to evaluate
  5. Check signs carefully: Multiplying by negative flips inequality!

🔥 The "Oscillation Check"

When you see these patterns, think SQUEEZE THEOREM:

  • \(\sin(1/x)\) or \(\cos(1/x)\) — oscillates infinitely fast near 0
  • Something bounded × something going to 0
  • Cannot use direct substitution or algebraic manipulation
  • Function "trapped" between two simpler functions

❌ Common Mistakes to Avoid

  • Mistake 1: Assuming \(\lim_{x \to 0} \sin(1/x)\) exists—it doesn't! The function oscillates
  • Mistake 2: Forgetting that BOTH bounds must have the SAME limit
  • Mistake 3: Reversing inequality when multiplying by negative number
  • Mistake 4: Not justifying the inequality on AP® exams
  • Mistake 5: Using \(f(a)\) instead of checking the squeeze—limit ≠ function value!
  • Mistake 6: Forgetting absolute value: \(-|x| \leq x\sin(1/x) \leq |x|\), not just \(-x\)
  • Mistake 7: Not recognizing when to use famous limits (\(\sin(x)/x = 1\))

🔄 Important Variations

Extensions of the Basic Limits

For any constant \(k\):

\[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \]
\[ \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \]
\[ \lim_{x \to 0} \frac{\tan(x)}{x} = 1 \]
\[ \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2} \]

✏️ AP® Exam Tips

What the AP® Exam Expects:

  • State the inequality: Write "Since \(-1 \leq \sin(x) \leq 1\)..."
  • Show the manipulation: Demonstrate how you get the bounding inequality
  • Evaluate BOTH bounds: Calculate both \(\lim g(x)\) and \(\lim h(x)\)
  • Verify they're equal: State "Since both limits equal \(L\)..."
  • Cite the theorem: Write "By the Squeeze Theorem, \(\lim f(x) = L\)"
  • Know the famous limits: \(\sin(x)/x\) and \((1-\cos(x))/x\) appear often!

FRQ Expectations (Topic 1.8):

  • Complete justification of inequalities
  • Correct evaluation of bounding limits
  • Proper notation and theorem citation
  • Connection to continuity (if functions are continuous)

⚡ Quick Reference Card

You See... Think... Use...
\(x\sin(1/x)\) Oscillating × vanishing \(-|x| \leq \cdots \leq |x|\)
\(x^2\cos(1/x)\) Bounded × vanishing squared \(-x^2 \leq \cdots \leq x^2\)
\(\sin(x)/x\) near 0 Famous limit! Result = 1
\((1-\cos(x))/x\) near 0 Famous limit! Result = 0
\(\sin(x)/x\) at \(\infty\) Bounded/unbounded \(-1/x \leq \cdots \leq 1/x\)

📝 Practice Problems

Try these yourself, then check the patterns:

  1. \(\lim_{x \to 0} x^3\sin(1/x)\)
  2. \(\lim_{x \to \infty} \frac{\cos(x)}{x}\)
  3. \(\lim_{x \to 0} \frac{\sin(7x)}{3x}\)
  4. If \(2x + 1 \leq f(x) \leq x^2 + 2x + 1\), find \(\lim_{x \to 0} f(x)\)

Answers:

  1. 0 (use \(-x^3 \leq x^3\sin(1/x) \leq x^3\))
  2. 0 (use \(-1/x \leq \cos(x)/x \leq 1/x\))
  3. \(7/3\) (rewrite as \((7/3) \cdot \sin(7x)/(7x)\))
  4. 1 (both bounds → 1)

🔗 Connection to Other Topics

Unit 1.8 (Squeeze Theorem) connects to:

  • Unit 1.7: Another tool in your limit-solving toolkit
  • Unit 2.4: Derivative of \(\sin(x)\) uses \(\lim \sin(x)/x = 1\)
  • Unit 3: Analyzing oscillating functions and their extrema
  • Unit 8: Convergence of series (squeeze test for sequences)
  • Throughout calculus: A powerful technique when algebraic methods fail

Remember: The Squeeze Theorem is like a mathematical vise—when you trap a function between two converging bounds, it has nowhere to go but toward the same limit! Master the two famous limits (\(\sin(x)/x = 1\) and \((1-\cos(x))/x = 0\)), recognize oscillating functions multiplied by vanishing terms, and always justify your inequalities. The squeeze is your secret weapon for limits that can't be evaluated any other way! 🥪✨