Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Additional and Advanced Exercises - Page 102: 3


Decrease, $0$.

Work Step by Step

Step 1. Given the equation $L=L_0 \sqrt {1-v^2/c^2}$, as $v$ increases, the term $v^2/c^2$ will also increase; however, the term $1-v^2/c^2$ will decrease, resulting in the decrease of $L$. Step 2. $\lim_{v\to c^-}L= L_0 \sqrt {1-c^2/c^2}=0$, the left-hand limit is needed because the speed of light is the maximum speed achievable by an object; in other words $v\leq c$.
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