Squeeze Theorem and Limits

Squaring Numbers and Inequalities

  • If you square a positive number, you get a positive number.
  • If you square a negative number, you get a positive number.

Multiplying Inequalities with Positive Numbers

  • When multiplying an inequality by a positive number, the sign does not flip.
  • Example: Multiplying an inequality by x^2 because x^2 is always positive.
    • We want to find the limit of x^2 \cdot cos(\frac{1}{x}) as x approaches 0.
    • To achieve this, multiply each side of the inequality by x^2.
    • x^2 \cdot (-1) \le x^2 \cdot cos(\frac{1}{x}) \le x^2 \cdot (1)
    • -x^2 \le x^2 \cdot cos(\frac{1}{x}) \le x^2
    • Here, f(x) = -x^2 and h(x) = x^2
  • Finding the limit:
    • \lim_{x \to 0} -x^2 = -0^2 = 0
    • \lim_{x \to 0} x^2 = 0^2 = 0
  • By the Squeeze Theorem, the limit of the function in the middle is also 0.
    • \lim_{x \to 0} x^2 \cdot cos(\frac{1}{x}) = 0

Squeeze Theorem Examples

  • Example 1: Show that \lim_{x \to 0} x^2 sin(20 \pi x) = 0 using the Squeeze Theorem.
  • Example 2: Show that \lim_{x \to 0} \sqrt{x^3 + x^2} sin(\frac{\pi}{x}) = 0 using the Squeeze Theorem.
  • Example 3: Evaluate \lim_{x \to 1} g(x) if 2x \le g(x) \le x^4 - x^2 + 2
  • Example 4: Find \lim_{x \to 4} f(x) if 4x - 9 \le f(x) \le x^2 - 4x + 7 for x \ge 0

Example 1 Solution: Finding the Limit of x^2 sin(20 \pi x)

  • Let g(x) = x^2 sin(20 \pi x)
  • We need to find functions f(x) and h(x) such that f(x) \le g(x) \le h(x) when x is near 0.
  • We know that -1 \le sin(20 \pi x) \le 1 because the sine function's range is always between -1 and 1.
  • Multiply all sides of the inequality by x^2, which is a positive number (since x \ne 0):
    • x^2 \cdot (-1) \le x^2 sin(20 \pi x) \le x^2 \cdot (1)
    • -x^2 \le x^2 sin(20 \pi x) \le x^2
  • Now, we have f(x) = -x^2 and h(x) = x^2
  • Evaluate the limits of f(x) and h(x) as x approaches 0:
    • \lim_{x \to 0} -x^2 = -0^2 = 0
    • \lim_{x \to 0} x^2 = 0^2 = 0
  • Since both limits are 0, by the Squeeze Theorem, the limit of g(x) is also 0:
    • \lim_{x \to 0} x^2 sin(20 \pi x) = 0

Example 2 Solution: Finding the Limit of \sqrt{x^3 + x^2} sin(\frac{\pi}{x})

  • We know that -1 \le sin(\frac{\pi}{x}) \le 1 for all x \ne 0
  • For \sqrt{x^3 + x^2} to be defined, x^3 + x^2 must be non-negative (i.e., positive or zero).
  • Multiply all sides of the inequality by \sqrt{x^3 + x^2}, which is non-negative:
    • -\sqrt{x^3 + x^2} \le \sqrt{x^3 + x^2} sin(\frac{\pi}{x}) \le \sqrt{x^3 + x^2}
  • Evaluate the limits of -\sqrt{x^3 + x^2} and \sqrt{x^3 + x^2} as x approaches 0:
    • \lim_{x \to 0} -\sqrt{x^3 + x^2} = -\sqrt{0^3 + 0^2} = 0
    • \lim_{x \to 0} \sqrt{x^3 + x^2} = \sqrt{0^3 + 0^2} = 0
  • Since both limits are 0, by the Squeeze (or Sandwich) Theorem:
    • \lim_{x \to 0} \sqrt{x^3 + x^2} sin(\frac{\pi}{x}) = 0

Math 107 Reminder:

  • If a \le b and c is a positive number, then c \cdot a \le c \cdot b and \frac{a}{c} \le \frac{b}{c}.
  • If c is a negative number, then a \cdot c \ge b \cdot c (the inequality flips).

Example 3 Solution: Evaluating \lim_{x \to 1} g(x)

  • Given 2x \le g(x) \le x^4 - x^2 + 2
  • Evaluate the limits of the bounding functions as x approaches 1:
    • \lim_{x \to 1} 2x = 2 \cdot 1 = 2
    • \lim_{x \to 1} (x^4 - x^2 + 2) = 1^4 - 1^2 + 2 = 1 - 1 + 2 = 2
  • Since both limits are equal to 2, by the Squeeze Theorem:
    • \lim_{x \to 1} g(x) = 2

Example 4 Solution: Finding \lim_{x \to 4} f(x)

  • Given 4x - 9 \le f(x) \le x^2 - 4x + 7
  • Find the limit of 4x-9 as x approaches 4:
    • \lim_{x \to 4} (4x - 9) = 4(4) - 9 = 16 - 9 = 7
  • Find the limit of x^2 - 4x + 7 as x approaches 4:
    • \lim_{x \to 4} (x^2 - 4x + 7) = 4^2 - 4(4) + 7 = 16 - 16 + 7 = 7
  • Since both limits are 7, by the Squeeze Theorem:
    • \lim_{x \to 4} f(x) = 7
  • Alternative representation:
    • \lim{x \to 4} (4x - 9) \le \lim{x \to 4} f(x) \le \lim_{x \to 4} (x^2 - 4x + 7)
    • 7 \le \lim_{x \to 4} f(x) \le 7
  • Therefore, the limit of f(x) as x approaches 4 is 7.