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 9th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1337613924
ISBN 13:
978-1-33761-392-7
Chapter 2 - Section 2.2 - The Limit of a Function - 2.2 Exercises - Page 94: 51
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
