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 - Page 97: 57

Answer

$$\eqalign{ & a.\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) = 1;\,\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = 1;\,\,\,\,\,\,y = 1 \cr & b.\,x = 0;\,\,\,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \infty ;\,\,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = + \infty \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{{x^2} - 9}}{{x\left( {x - 3} \right)}} \cr & {\text{Factor the numerator}} \cr & f\left( x \right) = \frac{{\left( {x + 3} \right)\left( {x - 3} \right)}}{{x\left( {x - 3} \right)}} \cr & f\left( x \right) = \frac{{x + 3}}{x} \cr & f\left( x \right) = 1 + \frac{3}{x} \cr & \cr & a.{\text{ Calculate }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr & *\,\,\mathop {\lim }\limits_{x \to \infty } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{3}{x}} \right) \cr & {\text{Use limits properties}} \cr & = \mathop {\lim }\limits_{x \to \infty } \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {\frac{3}{x}} \right)}^{{\text{approaches to 0}}} \cr & = 1 + 0 \cr & = 1 \cr & and\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \cr & *\,\,\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {1 + \frac{3}{x}} \right) \cr & = \mathop {\lim }\limits_{x \to - \infty } \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{3}{x}} \right)}^{{\text{approaches to 0}}} \cr & \,\, = 1 + 0 \cr & \,\, = 1 \cr & {\text{Then the horizontal asymptote of }}f\left( x \right){\text{ is }}y = 1 \cr & \cr & b.\,\,{\text{The function is not defined for }}x = 0,{\text{ then}} \cr & \,\,\,\,\,{\text{there is a vertical asymptotes at }}x = 0 \cr & \cr & {\text{Analyze }}\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) \cr & \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {1 + \frac{3}{x}} \right) \cr & {\text{ }} = \mathop {\lim }\limits_{x \to {0^ - }} \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to {0^ - }} \left( {\frac{3}{x}} \right)}^{{\text{approaches to - }}\infty } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \frac{3}{{{0^ - }}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - \infty \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \infty \cr & \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {1 + \frac{3}{x}} \right) \cr & {\text{ }} = \mathop {\lim }\limits_{x \to {0^ + }} \left( 1 \right) + \overbrace {\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{3}{x}} \right)}^{{\text{approaches to + }}\infty } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \infty \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \infty \cr} $$
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