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.
 A: 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.
A: 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 ($\sqrt{n^2+3}+ \sqrt{n^2 - 1} \sim n + n = 2n$)
A: $$\lim_{n\to \infty}(\sqrt[m]{an^m+bn^{n-1}+ ...}) \sim \sqrt[m]{a}.(n + \frac{b}{ma})$$
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1})) = \lim_{n\to \infty}(\sqrt{n^4+3n^2}-\sqrt{n^4-n^2}) = \lim_{t\to \infty}(\sqrt{t^2+3t}-\sqrt{t^2-t})\\ = \sqrt[2]{1}.(t + \frac{3}{2}) - \sqrt[2]{1}.(t - \frac{1}{2}) =\frac{3}{2} + \frac{1}{2} = 2$$
* Don't use this method if you have enough time to solve problems and enjoy them!
A: Things will be more clear if we set $\displaystyle n=\frac1h$
$$\lim_{n\to \infty}(n(\sqrt{n^2+3}-\sqrt{n^2-1}))=\lim_{h\to0^+}\frac{\sqrt{1+3h^2}-\sqrt{1-h^2}}{h^2}$$
$$=\lim_{h\to0^+}\frac{(1+3h^2)-(1-h^2)}{h^2(\sqrt{1+3h^2}+\sqrt{1-h^2})}\text{ (Rationalizing the numerator)}$$
$$=\lim_{h\to0^+}\frac{4h^2}{h^2}\cdot\frac1{\lim_{h\to0^+}(\sqrt{1+3h^2}+\sqrt{1-h^2})}$$
As $h\to0,h\ne0$ so the limit will be $$=\frac4{\sqrt1+\sqrt1}=\cdots$$
