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 - Page 138: 20


$$\lim\limits_{t\to\infty}\frac{t-t\sqrt t}{2t^{3/2}+3t-5}=\frac{-1}{2}$$

Work Step by Step

$$A=\lim\limits_{t\to\infty}\frac{t-t\sqrt t}{2t^{3/2}+3t-5}$$$$A=\lim\limits_{t\to\infty}\frac{t-t\sqrt t}{(2\times t\times t^{1/2})+3t-5}$$$$A=\lim\limits_{t\to\infty}\frac{t-t\sqrt t}{2t\sqrt t+3t-5}$$ Divide both numerator and denominator by $t\sqrt t$ We also have $t\sqrt t=t\times t^{1/2}=t^{3/2}$$$A=\lim\limits_{t\to\infty}\frac{\frac{t}{t\sqrt t}-\frac{t\sqrt t}{t\sqrt t}}{\frac{2t\sqrt t}{t\sqrt t}+\frac{3t}{t\sqrt t}-\frac{5}{t\sqrt t}}$$$$A=\lim\limits_{t\to\infty}\frac{\frac{1}{\sqrt t}-1}{2+\frac{3}{\sqrt t}-\frac{5}{t\sqrt t}}$$$$A=\lim\limits_{t\to\infty}\frac{\frac{1}{t^{1/2}}-1}{2+\frac{3}{t^{1/2}}-\frac{5}{t^{3/2}}}$$$$A=\frac{0-1}{2+3\times0-5\times0}$$$$A=\frac{-1}{2}$$
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