Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 323: 61

Answer

$y = x-1~~$ is a slant asymptote.

Work Step by Step

$y = \frac{x^2+1}{x+1}$ We can express $x^2+1$ in terms of $x+1$: $x^2+1 = (x+1)(x-1)+2$ Then: $\frac{x^2+1}{x+1} = (x-1)+\frac{2}{x+1}$ We can see that $~~y = x-1~~$ is a good candidate for a slant asymptote. We can evaluate the limit as $x \to -\infty$: $\lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-(x-1)]$ $= \lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-\frac{(x-1)(x+1)}{x+1}]$ $= \lim\limits_{x \to -\infty} (\frac{x^2+1}{x+1}-\frac{x^2-1}{x+1})$ $= \lim\limits_{x \to -\infty} \frac{2}{x+1}$ $= 0$ We can evaluate the limit as $x \to \infty$: $\lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-(x-1)]$ $= \lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-\frac{(x-1)(x+1)}{x+1}]$ $= \lim\limits_{x \to -\infty} (\frac{x^2+1}{x+1}-\frac{x^2-1}{x+1})$ $= \lim\limits_{x \to -\infty} \frac{2}{x+1}$ $= 0$ Therefore, $~~y = x-1~~$ is a slant asymptote.
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