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: 61

Answer

Converges

Work Step by Step

Apply the limit comparison test with $e^{-x/2}$ $\lim\limits_{x \to \infty} \dfrac{\dfrac{1}{\sqrt {e^x-x}}}{1/\sqrt {e^x}}=\lim\limits_{x \to \infty} \dfrac{\sqrt {e^x}}{\sqrt {e^x-x}}$ Apply L'Hopital's Rule: $\sqrt {\lim\limits_{x \to \infty} \dfrac{e^x}{e^x-1}} = \sqrt {\lim\limits_{x \to \infty} \dfrac{e^x}{e^x}}=\sqrt {1}=1$ Now, $\int_1^{\infty} e^{-x/2} dx=\lim\limits_{a \to \infty}\int_1^{a} e^{-x/2} dx \\=\lim\limits_{a \to \infty}[-2e^{-x/2}]_1^a \\=\lim\limits_{a \to \infty}[-2e^{-a/2}-(-2e^{-1/2})] \\=\dfrac{2}{\sqrt e}$ Thus, the integral converges.
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