How do I solve this $\lim\limits_{s\to \infty}\frac{2s^2}{s^2-4}$ How do I solve this  $$\lim_{s\to \infty}\frac{2s^2}{s^2-4}$$ .
I have tried partial fraction, but it does not give me a constant. And I know the answer is 2.

the answer is $2$. Note that $\lim\limits_{s \to \infty} \frac{M}{s^p} = 0 $ for any $M$ finite and $p\geq 1$.
$$ \lim_{s \to \infty} \frac{2s^2}{s^2 - 4} = \lim_{s \to \infty} \frac{2}{1 - 4/s^2} = \frac{2}{\lim\limits_{s \to \infty } 1 - \frac{4}{s^2} } = \frac{2}{1} = 2 $$
