<s>[INST] 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

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

 [/INST] 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
$$
-\frac 12\frac{\sqrt 5}{5}
$$
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