Speed of light, and limits

Question

From the theory of relativity there is this equation:

$$m=\frac{m_0}{\sqrt{1-(\frac{v^2}{c^2}})}$$

The problem asks to find the mass of the an object as it approaches the speed of light. I know the answer is $+\infty$, but I don't know how to write the limit. Would I just assume its like a regular function?

Assumption

$$f(v)=\frac{m_0}{\sqrt{1-(\frac{v^2}{c^2}})}$$ $$\lim_\limits{v \to c^-}(\frac{m_0}{\sqrt{1-(\frac{v^2}{c^2})}})$$ $$m_0 \lim_\limits{v \to c^-}((\sqrt{1-\frac{v^2}{c^2}})^{-1})$$

I just typically get up to there, and state that the mass, $m$ would approach $+\infty$. Are my assumptions correct, as well as my answer?

Answer 1

I think the most intuitive way to do this is in pieces.

As $v\rightarrow c^-$, we get $\frac{v^2}{c^2}\rightarrow 1^-$, so $1-\frac{v^2}{c^2}\rightarrow 0^+$, $\sqrt{1-\frac{v^2}{c^2}}\rightarrow 0^+$, and finally $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\rightarrow +\infty$.

Answer 2

If $f $ is continuous and $\lim_{x\rightarrow a} x $ exists then $\lim_{x\rightarrow a} f (x)=f (\lim x) $

So yes, you can do nearly all you want to do.

$\lim_{v\rightarrow c} \frac m {\sqrt {1-\frac {v^2}{c^2}}}=m\sqrt {\lim_v \lim_{x\rightarrow 1-v^2/c^2}(1/x)} $.

Which is infinite.