Xét \(m(v) = \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
Tập xác định: \(D = \mathbb{N}\backslash \{ c\} \)
Ta có: \(\mathop {\lim }\limits_{v \to {c^ + }} m(v) = \mathop {\lim }\limits_{v \to {c^ + }} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \mathop {\lim }\limits_{v \to {c^ + }} \frac{{\frac{{{m_0}}}{v}}}{{\sqrt {\frac{1}{{{v^2}}} - \frac{1}{{{c^2}}}} }} = \frac{{\frac{{{m_0}}}{c}}}{{\sqrt {\frac{1}{{{c^2}}} - \frac{1}{{{c^2}}}} }} = + \infty \); \(\mathop {\lim }\limits_{v \to {c^ - }} m(v) = \mathop {\lim }\limits_{v \to {c^ - }} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} = \mathop {\lim }\limits_{v \to {c^ - }} \frac{{\frac{{{m_0}}}{v}}}{{\sqrt {\frac{1}{{{v^2}}} - \frac{1}{{{c^2}}}} }} = \frac{{\frac{{{m_0}}}{c}}}{{\sqrt {\frac{1}{{{c^2}}} - \frac{1}{{{c^2}}}} }} = - \infty \)
Vậy đường thẳng x = c là tiệm cận đứng của đồ thị hàm số