Calculus: Early Transcendentals (2nd Edition)

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

Chapter 2 - Limits - Review Exercises - Page 124: 35

Answer

$$\frac{{3\pi }}{2} + 2$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to \infty } \left( {3{{\tan }^{ - 1}}x + 2} \right) \cr & {\text{Apply the property }}\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right) \cr & = \mathop {\lim }\limits_{x \to \infty } \left( {3{{\tan }^{ - 1}}x} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 2 \right) \cr & {\text{Apply the property }}\mathop {\lim }\limits_{x \to a} \left[ {kf\left( x \right)} \right] = k\mathop {\lim }\limits_{x \to a} \left[ {f\left( x \right)} \right] \cr & = 3\mathop {\lim }\limits_{x \to \infty } \left( {{{\tan }^{ - 1}}x} \right) + \mathop {\lim }\limits_{x \to \infty } \left( 2 \right) \cr & {\text{Where }}\mathop {\lim }\limits_{x \to \infty } \left( {{{\tan }^{ - 1}}x} \right) = \frac{\pi }{2} \cr & = 3\left( {\frac{\pi }{2}} \right) + 2 \cr & = \frac{{3\pi }}{2} + 2 \cr} $$
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