Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises: 54

Answer

As $v\rightarrow c^-, m\rightarrow \infty$

Work Step by Step

As $v\rightarrow c^-,$ $m\rightarrow \frac{m_0}{\sqrt{1-\frac{(c^-)^2}{c^2}}}=\frac{m_0}{\sqrt{1-\frac{(c^2)^-}{c^2}}}=\frac{m_0}{\sqrt{1-1^-}}=\frac{m_0}{0^+}=\infty$
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