Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises - Page 138: 28

Answer

$\lim\limits_{q \to \infty}\frac{q^3+6q-4}{4q^2-3q+3}=\infty$

Work Step by Step

$\lim\limits_{q \to \infty}\frac{q^3+6q-4}{4q^2-3q+3}=\lim\limits_{q \to \infty}\frac{q^3+6q-4}{4q^2-3q+3}\times \frac{1/q^2}{1/q^2}$ $=\lim\limits_{q \to \infty}\frac{q+6/q-4/q^2}{4-3/q+3/q^2}$ (Use the properties of limits) $=\frac{\lim\limits_{q \to \infty}q+\lim\limits_{q \to \infty}6/q-\lim\limits_{q \to \infty}4/q^2}{\lim\limits_{q \to \infty}(4-3/q+3/q^2)}$ (Use the limits $\lim\limits_{q \to \infty}q=\infty$ and $\lim\limits_{q \to \infty}k/q^n=0$ for $n=1,2,...$) $=\frac{\infty+0-0}{4-0+0}$ $=\frac{\infty}{4}$ $=\infty$ Thus, $\lim\limits_{q \to \infty}\frac{q^3+6q-4}{4q^2-3q+3}=\infty$.
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