Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 47

Answer

Vertical: $x = -3$ Horizontal: $y = 4$
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Work Step by Step

To find the horizontal asymptote, we need to find: $\lim\limits_{x \to \infty} \frac{5 + 4x}{x+3}$ and $\lim\limits_{x \to -\infty} \frac{5 + 4x}{x+3}$ First we divide both numerator and denominator by $x$. $\lim\limits_{x \to \infty} \frac{\frac{5+4x}{x}}{\frac{x+3}{x}} =$ $\lim\limits_{x \to \infty} \frac{\frac{5}{x}+4}{\frac{3}{x}+1} = \frac{0+4}{1+0} = 4$ Now with $-\infty$. $\lim\limits_{x \to -\infty} \frac{\frac{5+4x}{x}}{\frac{x+3}{x}} =$ $\lim\limits_{x \to -\infty} \frac{\frac{5}{x}+4}{\frac{3}{x}+1} = \frac{0+4}{1+0} = 4$ So with both answers we know that the horizontal asymptote is $y = 4$. To find the vertical asymptote we take the denominator $x+3$ and compare it to zero. $x + 3 = 0$ $x = -3$ So the vertical asymptote is $x = -3$ and the horizontal asymptote is $y = 4$.
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