<s>[INST] 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.
 [/INST] For positive $a$:
$$
\sqrt{a^2 + 4} = a\sqrt{1+\frac{4}{a^2}}
$$
and
$$\lim_{a\to\infty}\sqrt{1+\frac{4}{a^2}} = 1.
$$
So, the Taylor series of $\sqrt{1+x}$ around $0$ (when $x\to 0$) is:
$$
\sqrt{1+x} = 1 + \left.\frac{1}{2\sqrt{1+x}}\right\rvert_{x=0} x + O(x^2) = 1+ \frac{x}{2} + O(x^2)
$$
So, when $1<<a$ (it's important for $a$ to be big!) and putting $x = \frac{4}{a^2}$
$$
\sqrt{a^2 + 4} = a\left(1 + \frac{4}{2a^2} + O\left(\frac{1}{a^4}\right)\right)=a + \frac{2}{a} + O\left(\frac{1}{a^3}\right)
$$
EDIT: Corrected my answer.
</s>