In the theory of relativity, the mass of a particle with velocity v is m=m01−v2c2...

${\displaystyle \underset{v\to c^{-}}{lim}\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\mathrm{\infty }}$
Since v is close to c and ${\displaystyle v<c\to v^2<c^2\to \frac{v^2}{c^2}<1}$
The direction of the inequality does not change because v and c are the speed of the velocity of the particle and the velocity of light, respectively. The magnitude of a vector is always a positive number.
${\displaystyle -\frac{v^2}{c^2}\succ 1}$ (multiplying both sides by -1)
${\displaystyle 1-\frac{v^2}{c^2}>0\to }$ the denominator is a small and positive number.
The numerator ${\displaystyle m_0}$ is a positive number
The division is then a huge positive number.