Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 854: 5

Answer

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

Work Step by Step

Divide numerator and denominator with $t^{2}$ and we have $\lim\limits_{t \to \infty}\dfrac{1+t^{2}}{1-t^{2}}=\lim\limits_{t \to \infty}\dfrac{(1/t^{2})+1}{(1/t^{2})-1}=\dfrac{0+1}{0-1}=-1$ $\lim\limits_{t \to \infty} \tan^{-1}(t) =\dfrac{\pi}{2}$ and $\lim\limits_{t \to \infty} \dfrac{1-e^{-2t}}{t}=\lim\limits_{t \to \infty}\dfrac{1}{t}-\dfrac{1}{te^{2t}}=0$ Hence, we have $\displaystyle \lim_{t\rightarrow\infty}\lt \dfrac{1+t^{2}}{1-t^{2}},\ \tan^{-1}t,\dfrac{1-e^{-2t}}{t} \gt=\lt -1,\ \dfrac{\pi}{2},0 \gt$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.