How do I solve this $\lim\limits_{s\to \infty}\frac{2s^2}{s^2-4}$

Question

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.

Answer 1

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

Answer 2

Hint : $$\frac{2s^2}{s^2-4} = \frac{2}{1- \frac{4}{s^2}}$$

Answer 3

Since $s \to \infty$ we got a case for the rule of l'Hospital. Using this we get:

$$\frac{2}{2s}, s \to \infty $$ $$ \lim_{s\to \infty} \frac{1}{s}= 0$$

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Edit:

Still l'Hospital's rule can be applied $$\lim_{s\to \infty} \frac{2s^2}{s^2 -4}\stackrel{\text{l'hosp.}}{=}\frac{4s}{2s} = \frac{4}{2} = 2$$

Answer 4

As an alternative

$$\frac{2s^2}{s^2-4}=2+\frac{8}{s^2-4} \to 2+0=2$$

Answer 5

$$L=\lim_{s\to \infty}\frac{2s^2}{s^2-4}$$ Using fraction decomposition: $$=\lim_{s\to \infty}\frac{2s^2-8+8}{s^2-4}$$ $$=\lim_{s\to \infty}2+\lim_{s\to \infty}\frac{8}{s^2-4}$$ $$\implies L=2$$