AP Calculus BC 1.8 Determining Limits Using the Squeeze Theorem Study Notes - New Syllabus
AP Calculus BC 1.8 Determining Limits Using the Squeeze Theorem Study Notes- New syllabus
AP Calculus BC 1.8 Determining Limits Using the Squeeze Theorem Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
Determine the limits of functions using equivalent expressions for the function or the squeeze theorem.
Key Concepts:
- Using the Squeeze Theorem to evaluate limits
Determining Limits Using the Squeeze Theorem
Determining Limits Using the Squeeze Theorem
The Squeeze Theorem is a method for finding the limit of a function by comparing it to two other functions that “squeeze” it from above and below.
Theorem Statement:
If \( g(x) \le f(x) \le h(x) \) for all \( x \) near \( a \) (except possibly at \( a \)) and
\( \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L, \)
then \( \lim_{x \to a} f(x) = L \).
When to Use:
- When the function is complicated but bounded between two simpler functions.
- Especially useful for trigonometric limits near zero.
Common Example:
\( -1 \le \sin x \le 1 \implies -|x| \le x\sin\left(\frac{1}{x}\right) \le |x|. \)
Example:
Find \( \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) \).
▶️ Answer/Explanation
We know \( -1 \le \sin\left(\frac{1}{x}\right) \le 1 \).
Multiply by \( x \): \( -|x| \le x \sin\left(\frac{1}{x}\right) \le |x| \).
As \( x \to 0 \), both \( -|x| \) and \( |x| \) go to 0.
By the Squeeze Theorem, \( \lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0 \).
Example :
Compute \( \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) \).
▶️ Answer/Explanation
Since \( -1 \le \cos\left(\frac{1}{x}\right) \le 1 \), multiplying by \( x^2 \ge 0 \):
\( -x^2 \le x^2 \cos\left(\frac{1}{x}\right) \le x^2 \).
As \( x \to 0 \), both bounds \( \to 0 \).
Answer: \( 0 \).
Example :
Prove \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) using the Squeeze Theorem.
▶️ Answer/Explanation
For \( 0 < x < \frac{\pi}{2} \), we have \( \cos x \le \frac{\sin x}{x} \le 1 \).
As \( x \to 0^+ \), \( \cos x \to 1 \) and upper bound is \( 1 \).
By the Squeeze Theorem: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \).
Example:
Compute \( \lim_{x \to 0} x^2 \sin\left(\dfrac{1}{x}\right) \).
▶️ Answer/Explanation
Since \( -1 \le \sin\left(\dfrac{1}{x}\right) \le 1 \), multiply by \( x^2 \ge 0 \):
\( -x^2 \le x^2 \sin\left(\dfrac{1}{x}\right) \le x^2 \).
As \( x \to 0 \), both bounds \( \to 0 \).
By the Squeeze Theorem, \( \lim_{x \to 0} x^2 \sin\left(\dfrac{1}{x}\right) = 0 \).