real analysis - Calculate $lim_{nto{+}infty}{(sqrt{n^{2}+n}-n)}$

Could someone help me through this problem?

Calculate $\displaystyle\lim_{n \to{+}\infty}{(\sqrt{n^{2}+n}-n)}$

Answer

We have:

$$\sqrt{n^{2}+n}-n=\frac{(\sqrt{n^{2}+n}-n)(\sqrt{n^{2}+n}+n)}{\sqrt{n^{2}+n}+n}=\frac{n}{\sqrt{n^{2}+n}+n}$$
Therefore:

$$\sqrt{n^{2}+n}-n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$$

And since: $\lim\limits_{n\to +\infty}\frac{1}{n}=0$

It follows that:

$$\boxed{\,\,\lim\limits_{n\to +\infty}(\sqrt{n^{2}+n}-n)=\dfrac{1}{2}\,\,}$$