Answer
As $v\rightarrow c^-, m\rightarrow \infty$
Work Step by Step
As $v\rightarrow c^-,$
$m\rightarrow \frac{m_0}{\sqrt{1-\frac{(c^-)^2}{c^2}}}=\frac{m_0}{\sqrt{1-\frac{(c^2)^-}{c^2}}}=\frac{m_0}{\sqrt{1-1^-}}=\frac{m_0}{0^+}=\infty$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
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 - Page 94: 54
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