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 853: 3



Work Step by Step

Given: $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=(1)\mathrm{i}+(1)\mathrm{j}+(1) \mathrm{k}$ $ \lim\limits_{t\rightarrow 0} e^{-3t}=e^{0}=1$ and $\lim\limits_{t\rightarrow 0} \dfrac{t^{2}}{\sin^{2}t}=\lim\limits_{t\rightarrow 0} (\dfrac{\sin t}{t})^{-2}=[\lim\limits_{t\rightarrow 0} \dfrac{\sin t}{t}]^{-2}=1$ and $\lim\limits_{t\rightarrow 0} \cos 2t =\cos (0)=1$ $\lim\limits_{t\rightarrow 0} (e^{-3t}$i$+\dfrac{t^{2}}{\sin^{2}t}\mathrm{j}+\cos 2t \mathrm{k})=(1) \times \mathrm{i}+(1) \times \mathrm{j}+(1) \times \mathrm{k}=i+j+k$ Hence, we have $i+j+k$
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