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: 19

Answer

$\lim\limits_{t\to\infty}\frac{\sqrt t+t^2}{2t-t^2}=-1$

Work Step by Step

$$A=\lim\limits_{t\to\infty}\frac{\sqrt t+t^2}{2t-t^2}$$ Divide both numerator and denominator by $t^2$ $A=\lim\limits_{t\to\infty}\frac{\frac{\sqrt t}{t^2}+\frac{t^2}{t^2}}{\frac{2t}{t^2}-\frac{t^2}{t^2}}$ $A=\lim\limits_{t\to\infty}\frac{\sqrt\frac{t}{t^4}+1}{\frac{2}{t}-1}$ $A=\lim\limits_{t\to\infty}\frac{\sqrt\frac{1}{t^3}+1}{\frac{2}{t}-1}$ $A=\lim\limits_{t\to\infty}\frac{\frac{1}{t^{3/2}}+1}{\frac{2}{t}-1}$ $A=\frac{\lim\limits_{t\to\infty}(\frac{1}{t^{3/2}}+1)}{\lim\limits_{t\to\infty}(\frac{2}{t}-1)}$ $A=\frac{0+1}{2\times0-1}$ $A=-1$
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