A graphing calculator is recommended. Use the Squeeze Theorem to show that lim x² cos(167x) = 0. Illustrate by graphing the functions f(x) = -x², g(x) = x² cos(16x), and h(x) = x² on the same screen. ? ✓≤ x² сos(16πx) ≤ ? Let f(x) = -x², g(x) = x² cos(16x), and h(x) = x². Then ? ≤ cos(16x) ≤? V by the Squeeze Theorem we have lim g(x) lim f(x) = lim h(x) = = X-0 X-0 . Since
A graphing calculator is recommended. Use the Squeeze Theorem to show that lim x² cos(167x) = 0. Illustrate by graphing the functions f(x) = -x², g(x) = x² cos(16x), and h(x) = x² on the same screen. ? ✓≤ x² сos(16πx) ≤ ? Let f(x) = -x², g(x) = x² cos(16x), and h(x) = x². Then ? ≤ cos(16x) ≤? V by the Squeeze Theorem we have lim g(x) lim f(x) = lim h(x) = = X-0 X-0 . Since
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
![### Understanding Limits and the Squeeze Theorem
A graphing calculator is recommended.
#### Use the Squeeze Theorem to show that:
\[ \lim_{x \to 0} x^2 \cos(16\pi x) = 0 \]
#### Illustrate by graphing the functions \( f(x) = -x^2 \), \( g(x) = x^2 \cos(16\pi x) \), and \( h(x) = x^2 \) on the same screen.
Let:
- \( f(x) = -x^2 \)
- \( g(x) = x^2 \cos(16\pi x) \)
- \( h(x) = x^2 \)
Then, using the Squeeze Theorem:
\[ -1 \leq \cos(16\pi x) \leq 1 \]
\[ \Rightarrow -x^2 \leq x^2 \cos(16\pi x) \leq x^2 \]
Since:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} h(x) = 0 \]
By the Squeeze Theorem we have:
\[ \lim_{x \to 0} g(x) = 0 \]
Fill in the blanks:
- In the inequality \( [\boxed{-1}] \leq \cos(16\pi x) \leq [\boxed{+1}] \)
- In the inequality \( -x^2 \leq x^2 \cos(16\pi x) \leq x^2 \)
- In the final result: by the Squeeze Theorem, we have \( \lim_{x \to 0} x^2 \cos(16\pi x) = [\boxed{0}] \)
Thus, the Squeeze Theorem effectively shows that the limit of \( x^2 \cos(16\pi x) \) as \( x \) approaches 0 is 0.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00067d91-5c79-4b53-8e03-40299a52da99%2F1d9cbad6-dbbf-44c7-bcbd-9b5b85a6811a%2Fvzfw4vb_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Limits and the Squeeze Theorem
A graphing calculator is recommended.
#### Use the Squeeze Theorem to show that:
\[ \lim_{x \to 0} x^2 \cos(16\pi x) = 0 \]
#### Illustrate by graphing the functions \( f(x) = -x^2 \), \( g(x) = x^2 \cos(16\pi x) \), and \( h(x) = x^2 \) on the same screen.
Let:
- \( f(x) = -x^2 \)
- \( g(x) = x^2 \cos(16\pi x) \)
- \( h(x) = x^2 \)
Then, using the Squeeze Theorem:
\[ -1 \leq \cos(16\pi x) \leq 1 \]
\[ \Rightarrow -x^2 \leq x^2 \cos(16\pi x) \leq x^2 \]
Since:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} h(x) = 0 \]
By the Squeeze Theorem we have:
\[ \lim_{x \to 0} g(x) = 0 \]
Fill in the blanks:
- In the inequality \( [\boxed{-1}] \leq \cos(16\pi x) \leq [\boxed{+1}] \)
- In the inequality \( -x^2 \leq x^2 \cos(16\pi x) \leq x^2 \)
- In the final result: by the Squeeze Theorem, we have \( \lim_{x \to 0} x^2 \cos(16\pi x) = [\boxed{0}] \)
Thus, the Squeeze Theorem effectively shows that the limit of \( x^2 \cos(16\pi x) \) as \( x \) approaches 0 is 0.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning