<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] 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
$$\lim_{h\to0^+}\dfrac{\sqrt{5+4h^2}-\sqrt{5+h}}h=\lim_{h\to0^+}\dfrac1{\sqrt{5+4h^2}+\sqrt{5+h}}\cdot\lim_{h\to0^+}\dfrac{5+4h^2-(5+h)}h=?$$
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