## Calculus: Early Transcendentals 8th Edition

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