Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$ 
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$

Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.

My own solution

$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}~\frac{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}$$
$$\lim_{n\to\infty} \frac{5n^2+4~-~(5n^2+n)}{\sqrt{5n^2+4}~+~\sqrt{5n^2+n}}~=~\lim_{n\to\infty} \frac{4-n}{n\left(\sqrt{5+\frac4{n^2}}~+~\sqrt{5+\frac1{n}}\right)}$$
$$=~\frac{-1}{2\sqrt{5}}~=~-\frac{\sqrt{5}}{10}$$

 A: Let $1/n=h$
As $n\to\infty,h\to0+,h>0$
$$\sqrt{5n^2+4}=\sqrt{\dfrac{5+4h^2}{h^2}}=\dfrac{\sqrt{5+4h^2}}{\sqrt{h^2}}$$
Now $\sqrt{h^2}=|h|=+h$ for $h>0$
So, we ahve
$$\lim_{h\to0^+}\dfrac{\sqrt{5+4h^2}-\sqrt{5+h}}h=\lim_{h\to0^+}\dfrac1{\sqrt{5+4h^2}+\sqrt{5+h}}\cdot\lim_{h\to0^+}\dfrac{5+4h^2-(5+h)}h=?$$
A: Two alternative ideas:
(1) Write
$$
\sqrt{5n^2+4} - \sqrt{5n^2+n} =
\sqrt 5n(\sqrt{1 + 4/5n^2} - \sqrt{1 + 1/5n})
$$
and use Taylor ($\sqrt{1 + x} = 1 + x/2 - x^2/8 + \cdots$).
(2) Let be $f(x) = \sqrt x$. By the Mean Value Theorem, for some $c_n\in(5n^2 + 4,5n^2 + n)$
$$
\sqrt{5n^2 + 4} - \sqrt{5n^2 + n} =
f'(c_n)((5n^2 + 4) - (5n^2 + n)) =
-\frac{4 - n}{2\sqrt{c_n}} =
-\frac{4/n - 1}{2\sqrt{c_n/n^2}},
$$
and by sqeezing $c_n/n^2\to 5$.
A: Considering that $\frac{4}{n^2}<< \frac{1}{n}$ for big $n$ we have
$$
\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n} \equiv \lim_{h\to 0}\frac{\sqrt{5+4h^2}-\sqrt{5+h}}{h} = \lim_{h\to 0}\frac{\sqrt{5+o(h^2)}-\sqrt{5+h}}{h}
$$
now remember the derivative definition
$$
\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h} = \frac 12\frac{\sqrt x}{x}
$$
so the result is
$$
-\frac 12\frac{\sqrt 5}{5}
$$
