Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - Review Exercises - Page 331: 48

Answer

$$0$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \frac{{\ln {x^{100}}}}{{\sqrt x }} \cr & {\text{Evaluate }} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\ln {x^{100}}}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln {{\left( \infty \right)}^{100}}}}{{\sqrt \infty }} = \frac{\infty }{\infty } \cr & {\text{Apply the LHopital's rule}} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{\ln {x^{100}}}}{{\sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{d}{{dx}}\left[ {\ln {x^{100}}} \right]}}{{\frac{d}{{dx}}\left[ {\sqrt x } \right]}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{100}}{x}}}{{\frac{1}{{2\sqrt x }}}} \cr & = \mathop {\lim }\limits_{x \to \infty } \frac{{200\sqrt x }}{x} = \mathop {\lim }\limits_{x \to \infty } \frac{{200}}{{{x^{1/2}}}} \cr & {\text{Evaluate }} \cr & \mathop {\lim }\limits_{x \to \infty } \frac{{200}}{{{x^{1/2}}}} = \frac{{200}}{{{{\left( \infty \right)}^{1/2}}}} = 0 \cr} $$
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