Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.6 - Limits Involving Infinity; Asymptotes of Graphs - Exercises 2.6 - Page 98: 76

Answer

Sample answer: $\quad k(x)= 1+\displaystyle \frac{1}{1-x}$

Work Step by Step

Horizontal asymptote is $y=1$. There is a vertical asymptote, $x=1.$ $k(x)$ can have the form $k(x)=1+$(rational expression with degree of numerator less than that of the denominator) $(x-1)$ should be a factor of the denominator. $\displaystyle \frac{1}{x-1} \rightarrow-\infty$ when $x\rightarrow 1^{-}$, so we change the sign: $\displaystyle \frac{1}{1-x} \rightarrow+\infty$ when $x\rightarrow 1^{-}$, and $\displaystyle \frac{1}{1-x} \rightarrow-\infty$ when $x\rightarrow 1^{+}.$ $k(x)= 1+\displaystyle \frac{1}{1-x}$ satisfies the conditions of the problem.
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