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.

For the limit: We take advantage of obtaining a difference of squares. 
We have a factor of the form $a - b$, so we multiply it by $\dfrac{a+b}{a+b}$ to get $\dfrac{a^2 - b^2}{a+b}.$
Here, we multiply by $$\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$
$$n(\sqrt{n^2+3}-\sqrt{n^2-1})\cdot\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}} = \dfrac{n[n^2 + 3 - (n^2 - 1)]}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$
Now simplify and evaluate.
