Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 3 - Determining Change: Derivatives - 3.7 Activities - Page 244: 17


Indeterminate form is $\dfrac{\infty}{\infty}$ and$$\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1 } = \frac{3 }{5 }$$

Work Step by Step

Given $$\lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1}$$ using the method of replacement \begin{align*} \lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1 }&=\frac{\infty}{\infty} \end{align*} This is an indeterminate form, then applying L'Hôpital's Rule, we get \begin{align*} \lim _{x \rightarrow \infty} \frac{3 x^{2}+2 x+4}{5 x^{2}+x+1 }&=\lim _{x \rightarrow \infty} \frac{6 x+2 }{10x+1 } \ \ \text{apply L'Hôpital's Rule}\\ &=\lim _{x \rightarrow \infty} \frac{6 }{10 }\\ &= \frac{6 }{10 }= \frac{3 }{5 } \end{align*}
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