Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 894: 6


$\lt 0, \dfrac{1}{2},1 \gt$

Work Step by Step

Given: $\lim\limits_{t \to \infty} \lt te^{t}, \dfrac{t^3+t}{2t^3-1}, t \sin \dfrac{1}{t} \gt$ $\lt \lim\limits_{t \to \infty} te^{t}, \lim\limits_{t \to \infty} \dfrac{t^3+t}{2t^3-1}, \lim\limits_{t \to \infty} t \sin \dfrac{1}{t} \gt$ Need to apply L-Hospital's Rule. $\lt \lim\limits_{t \to \infty} te^{t}, \lim\limits_{t \to \infty} \dfrac{t^3+t}{2t^3-1}, \lim\limits_{t \to \infty} t \sin \dfrac{1}{t} \gt$ This implies that $\lt \lim\limits_{t \to \infty} \dfrac{t}{e^t}, \lim\limits_{t \to \infty} \dfrac{1+1/t^2}{2+1/t^3}, \lim\limits_{t \to \infty} \dfrac{-1/t^2 \cos (1/t)}{-1/t^2} \gt$ This gives: $\lt 0, \dfrac{1+0}{2+0}, \lim\limits_{t \to \infty} \cos (\dfrac{1}{t}) \gt$ Answer: $\lt 0, \dfrac{1}{2},1 \gt$
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