Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.6 - Limits at Infinity; Horizontal Asymptotes - 2.6 Exercises: 26

Answer

$$\lim\limits_{x\to\infty}\frac{x+3x^2}{4x-1}=\infty$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{x+3x^2}{4x-1}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x$, we have - In the numerator: $X=\frac{x+3x^2}{x}=1+3x$ - In the denominator: $Y=\frac{4x-1}{x}=4-\frac{1}{x}$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{1+3x}{4-\frac{1}{x}}$$ As $x\to\infty$, $(1+3x)\to\infty$ and $(4-\frac{1}{x})\to(4-0)=4$ Therefore, $A=\infty$
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