When $a\to \infty$, $\sqrt{a^2+4}$ behaves as $a+\frac{2}{a}$? What does it mean that $\,\,f(a)=\sqrt{a^2+4}\,\,$ behaves as $\,a+\dfrac{2}{a},\,$
as $a\to \infty$? 
How can this be justified? 
Thanks.

More precisely, $\sqrt{a^2+4} = a + 2/a + O(1/a^3)$ as $a \to \infty$.  In fact
it is easy to see that $$(a+2/a)^2 > a^2 + 4 > (a+2/a - 2/a^3)^2 \ \text{for} \ a > 1$$
so
$$a + \dfrac{2}{a} > \sqrt{a^2 + 4} > a + \dfrac{2}{a} - \dfrac{2}{a^3}$$
