University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 54



Work Step by Step

Apply the limit comparison test. Thus, we have $ \lim\limits_{x \to \infty} \dfrac{x/\sqrt{x^4-1}}{ x/\sqrt {x^{4}}}=\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{ \sqrt {x^4-1}}$ This suggests that $\lim\limits_{x \to \infty} \dfrac{\sqrt{x^4}}{ \sqrt {x^4-1}}=1$ Since we got a finite positive value, we know that both series converge or both diverge. This implies that $\int_2^{\infty} \dfrac{x dx}{\sqrt {x^4}}=\lim\limits_{p \to \infty}\int_2^{p} \dfrac{dx}{x}$ and $\lim\limits_{p \to \infty}\int_2^{p} \dfrac{dx}{x}=\lim\limits_{p \to \infty}[\ln|x|]_2^{p} =\infty$ Hence, the given integral diverges by the limit comparison test.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.