13. Let f (r) = a² cos² (4)(a) Use the squeeze play theorem to show that lim f (r) = 0.Hint. Observe that f (x) is never negative and use the fact that the cosine function hasa range between -1 and 1.(b) Show that f is discontinuous at c = 0 but is continuous at all other values.

The function is fx=x2cos21x As fx is never negative, fx=x2cos21x≥0. We also have 0≤cos21x≤1, then…

Answered: Let f (r) = r² cos² (4) (a) Use the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Concept explainers

Question

13. Let f (r) = a² cos² (4)
(a) Use the squeeze play theorem to show that lim f (r) = 0.
Hint. Observe that f (x) is never negative and use the fact that the cosine function has
a range between -1 and 1.
(b) Show that f is discontinuous at c = 0 but is continuous at all other values.

Expert Solution

Solution:

The function is $f (x) = x^{2} \cos^{2} (\frac{1}{x})$

As $f (x)$ is never negative, $f (x) = x^{2} \cos^{2} (\frac{1}{x}) \geq 0$ .

We also have $0 \leq \cos^{2} (\frac{1}{x}) \leq 1$ , then $0 \leq x^{2} \cos^{2} (\frac{1}{x}) \leq x^{2}$ .

Then apply limits:

$0 \leq \lim_{x \to 0} x^{2} \cos^{2} (\frac{1}{x}) \leq \lim_{x \to 0} x^{2} 0 \leq \lim_{x \to 0} x^{2} \cos^{2} (\frac{1}{x}) \leq 0 \lim_{x \to 0} x^{2} \cos^{2} (\frac{1}{x}) = 0$

Hence, by squeeze theorem $\lim_{x \to 0} x^{2} \cos^{2} (\frac{1}{x}) = 0$ .

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Limits and Continuity$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions