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: 25

Answer

$$\lim\limits_{x\to\infty}\frac{\sqrt{x+3x^2}}{4x-1}=\frac{\sqrt3}{4}$$

Work Step by Step

$$A=\lim\limits_{x\to\infty}\frac{\sqrt{x+3x^2}}{4x-1}$$$$A=\lim\limits_{x\to\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x$, we have - In the numerator: $X=\frac{\sqrt{x+3x^2}}{x}=\frac{\sqrt{x+3x^2}}{\sqrt {x^2}}=\sqrt{\frac{x+3x^2}{x^2}}=\sqrt{\frac{1}{x}+3}$ - In the denominator: $Y=\frac{4x-1}{x}=4-\frac{1}{x}$ Therefore, $$A=\lim\limits_{x\to\infty}\frac{\sqrt{\frac{1}{x}+3}}{4-\frac{1}{x}}$$$$A=\frac{\lim\limits_{x\to\infty}\sqrt{\frac{1}{x}+3}}{4-\lim\limits_{x\to\infty}\frac{1}{x}}$$$$A=\frac{\sqrt{\lim\limits_{x\to\infty}(\frac{1}{x})+3}}{4-\lim\limits_{x\to\infty}\frac{1}{x}}$$$$A=\frac{\sqrt{0+3}}{4-0}$$$$A=\frac{\sqrt3}{4}$$
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