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 97: 18


(a) $9/2$ (b) $ 9/2$

Work Step by Step

(a) \begin{align*} \lim_{x\to \infty}h(x)&= \lim_{x\to \infty} \frac{9x^4+x}{2x^4+5x^2 -x+6}\\ &= \lim_{x\to \infty} \frac{9x^4/x^4+x/x^4}{2x^4/x^4+5x^2/x^4 -x/x^4+6/x^4}\\ &= \frac{ \lim_{x\to \infty}(9 )+ \lim_{x\to \infty}(1/x^3)}{ \lim_{x\to \infty}(2)+ \lim_{x\to \infty}(5/x^2) - \lim_{x\to \infty}(1/x^3)+ \lim_{x\to \infty}(6/x^4)}\\ &= \frac{9+0}{2+0-0+0}\\ &=\frac{9}{2} \end{align*} (b) \begin{align*} \lim_{x\to -\infty}h(x)&= \lim_{x\to- \infty} \frac{9x^4+x}{2x^4+5x^2 -x+6}\\ &= \lim_{x\to -\infty} \frac{9x^4/x^4+x/x^4}{2x^4/x^4+5x^2/x^4 -x/x^4+6/x^4}\\ &= \frac{ \lim_{x\to -\infty}(9 )+ \lim_{x\to -\infty}(1/x^3)}{ \lim_{x\to -\infty}(2)+ \lim_{x\to- \infty}(5/x^2) - \lim_{x\to- \infty}(1/x^3)+ \lim_{x\to -\infty}(6/x^4)}\\ &= \frac{9+0}{2+0-0+0}\\ &=\frac{9}{2} \end{align*}
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