Solution to the limit of a series I'm strugling with the following problem:
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})), n \in \mathbb{N}$$
Wolfram Alpha says the answer is 2, but I don't know to calculate the answer.
Any help is appreciated.

$$\lim_{n\to \infty}(\sqrt[m]{an^m+bn^{n-1}+ ...}) \sim \sqrt[m]{a}.(n + \frac{b}{ma})$$
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})) = \lim_{n\to \infty}(\sqrt{n^4+3n^2}-\sqrt{n^4-n^2}) = \lim_{t\to \infty}(\sqrt{t^2+3t}-\sqrt{t^2-t})\\ = \sqrt[2]{1}.(t + \frac{3}{2}) - \sqrt[2]{1}.(t - \frac{1}{2}) =\frac{3}{2} + \frac{1}{2} = 2$$
* Don't use this method if you have enough time to solve problems and enjoy them!
