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: 48

Answer

Diverges.

Work Step by Step

Use the limit comparison test. $\lim\limits_{x \to \infty} \dfrac{\dfrac{1}{\sqrt {x}}-1}{\dfrac{1}{\sqrt {x}}}=\lim\limits_{x \to \infty} \dfrac{\sqrt x}{\sqrt {x}-1}$ Use L'Hospital's Rule: $ \lim\limits_{x \to \infty} \dfrac{\dfrac{1}{2 \sqrt {x}}}{1/2\sqrt x}=1$ and $\lim\limits_{a \to \infty}\int_{4}^{a} x^{-(1/2)} dx=\lim\limits_{a \to \infty}[2 \sqrt x]_{4}^a\\=\lim\limits_{a \to \infty}[2\sqrt a-4] \\=\infty$ So, the integral diverges.
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