Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.7 Limits at Infinity - Exercises - Page 82: 12


$$\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}=-\infty $$

Work Step by Step

Given $$\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}$$ let $$ f(x) = \frac{9 x^{2}-2}{6-29 x} $$ Since, we have $$ f(\infty)= \frac{\ \infty-\infty }{6-\infty}=\frac{\infty}{\infty}$$ So, transform algebraically and cancel \begin{aligned}L&=\lim _{x \rightarrow \infty} \frac{9 x^{2}-2}{6-29 x}\\ &=\lim _{x \rightarrow \infty} \frac{ \frac{9x^{2}}{x}-\frac{2}{x}}{\frac{6}{x}-\frac{29 x}{x}}\\ &=\lim _{x \rightarrow \infty} \frac{ 9x-\frac{2}{x}}{\frac{6}{x}-29}\\ \end{aligned} Since we have: \begin{aligned} \lim _{x \rightarrow \infty} \frac{ 1}{x}=0\\ \end{aligned} So, we get: \begin{aligned} L&= \frac{ \infty-0}{0-29}\\ &= \frac{ \infty }{ -29}\\ &=-\infty \end{aligned}
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