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:
  • $\lim<em>{x \to 4} (4x - 9) \le \lim</em>{x \to 4} f(x) \le \lim_{x \to 4} (x^2 - 4x + 7)$
  • $7 \le \lim_{x \to 4} f(x) \le 7$
• Therefore, the limit of f(x) as x approaches 4 is 7.