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.

it may seem obvious to use the asymptotic notation to solve this limit.
$$\sqrt{n^2 + a} \sim_{\infty} n $$
But substituting yields $n(n - n) = 0$ which is not allowed by theory (it basically means that our approximation isn't good enough to decide the limit)
so rationalize the numerator by multiplying by $$\dfrac{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$ as pointed out by amWhy
This yields 
$$\dfrac{4n}{\sqrt{n^2+3}+ \sqrt{n^2 - 1}}$$ Now we can use the asymptotic that yields calculations - free $$\dfrac{4n}{2n} = 2$$ since ($\sqrt{n^2+3}+ \sqrt{n^2 - 1} \sim n + n = 2n$)
