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.
 A: This is correct:
$$
\sqrt{a^2+4}=a\sqrt{1+\frac{4}{a^2}}=a+a\left(\sqrt{1+\frac{4}{a^2}}-1\right)=a+a\frac{1+\frac{4}{a^2}-1}{\sqrt{1+\frac{4}{a^2}}+1}\\=a+\frac{4}{a}\cdot\frac{1}{\sqrt{1+\frac{4}{a^2}}+1}\approx a+\frac{2}{a}
$$
A: 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
$$a + \dfrac{2}{a} > \sqrt{a^2 + 4} > a + \dfrac{2}{a} - \dfrac{2}{a^3}$$
A: $$\sqrt{a^2+4}\sim a+\frac{1}{a}$$
as $a\to\infty$ if 
$$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=1$$ But $$\lim_{a\to\infty}\frac{\sqrt{a^2+4}}{a+\frac{1}{a}}=\lim_{a\to\infty}\frac{a\sqrt{a^2+4}}{a^2+1}=\lim_{a\to\infty}\frac{a^2\sqrt{1+4/a^2}}{a^2(1+1/a^2)}=1$$
A: 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.
A: This goes back to Newton: The binomial theorem doesn't only work for natural number exponents!  If $a > 2$, the following formula holds:
$$\sqrt{a^2+ 4} = (a^2 + 4)^{1/2} = \sum_{k=0}^\infty {{1/2}\choose k}(a^2)^{1/2-k}4^k$$ $$= a + (1/2)\frac{4}{a} + \frac{(1/2)(-1/2)}{2!} \frac{4^2}{a^3} + \cdots$$ $$= a + \frac{2}{a} - \frac{2}{a^3} + \cdots$$
Notice that when $a$ is very large, the later terms become less and less significant.  
