Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 2 - Limits - 2.5 Limits at Infinity - 2.5 Exercises: 13

Answer

$0$

Work Step by Step

$\dfrac {-1}{\sqrt {x}}\leq \dfrac {\cos x^{5}}{\sqrt {x}}\leq \dfrac {1}{\sqrt {x}}$ $\lim _{x\rightarrow \infty }\dfrac {-1}{\sqrt {x}}=\dfrac {-1}{x^{\frac {1}{2}}}=0$ $\lim _{x\rightarrow \infty }\dfrac {1}{\sqrt {x}}=\dfrac {1}{x^{\frac {1}{2}}}=0$ $\lim _{x\rightarrow \infty }0\leq \dfrac {\cos x^{5}}{\sqrt {x}}\leq 0\Rightarrow \lim _{x\rightarrow \infty }\dfrac {\cos x^{5}}{\sqrt {x}}=0$
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