Thomas' Calculus 13th Edition

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

Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 12

Answer

$\dfrac{-2}{3}$

Work Step by Step

Consider: $\lim\limits_{x \to \infty}f(x)=\lim\limits_{x \to \infty}\dfrac{x-8x^2}{12x^2+5x}$ Need to check that the limit has an indeterminate form. Thus, $f(\infty)=\dfrac{\infty}{\infty}$ Now, apply L-Hospital's rule such as: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$ This implies: $\lim\limits_{x \to \infty}\dfrac{1-16x}{24x+5}=\dfrac{\infty}{\infty}$ Again apply L-Hospital's rule. $\lim\limits_{x \to \infty}\dfrac{-16}{24}=\dfrac{-2}{3}$
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