How to be sure about the limit of a given sequence?

Question

For example: $$ \lim_{n\to\infty}(\sqrt{n^2+2n}-\sqrt{n^2-2n})=(\sqrt{n^2+2n}-\sqrt{n^2-2n})(\frac{\sqrt{n^2+2n}+\sqrt{n^2-2n}}{\sqrt{n^2+2n}+\sqrt{n^2-2n}})=\frac{4n}{\sqrt{n}(\sqrt{n+2}+\sqrt{n-2})}=\frac{\lim_{n\to\infty}4n=\infty}{\lim_{\sqrt{n}(\sqrt{n+2}+\sqrt{n-2})}=\infty}=>\sqrt{n}(\sqrt{n+2}+\sqrt{n-2})=O(n). $$

But actually the limit is 2. How could I know that I have to search for an atcual limit?

Answer 1

There is no way to know befeore, we need to calculate properly, for this particular case form here

$$...=\frac{4n}{\sqrt{n}(\sqrt{n+2}+\sqrt{n-2})}=\frac n {\sqrt{n}\sqrt{n}}\frac{4}{(\sqrt{1+2/n}+\sqrt{1-2/n})}\to 2$$

Answer 2

Well. You have \begin{align} \frac{4n}{\sqrt{n}(\sqrt{n+2}+\sqrt{n-2}} = \frac{4}{(\sqrt{1+2/n}+\sqrt{1-2/n)}}\rightarrow 2 \end{align}