Prove using the definition of a limit that $$\displaystyle{\lim_{n\to\infty}\underbrace{\sqrt{n^2+n}-n}_{s_n}}=\frac{1}{2}.$$
Proof:
Let $\epsilon>0$ and $n\in\mathbb{N}$, then $$\left|\sqrt{n^2+n}-n-\frac12\right|\stackrel{(*)}{=}\left|\frac{n-\sqrt{n^2+n}}{2(n+\sqrt{n^2+n})}\right|=\left|\frac{n}{2(n+\sqrt{n^2+n})^2}\right|=\left|\frac{n}{4n^2+4n\sqrt{n^2+n}+2n}\right|<\left|\frac{n}{4n\sqrt{n^2+n}+2n}\right|=\left|\frac{1}{4\sqrt{n^2+n}+2}\right|$$ Since $\forall n\in\mathbb{N}\,\,\sqrt{n^2+n}>\sqrt{n}\,$ then $$\left|\frac{1}{4\sqrt{n^2+n}+2}\right|<\left|\frac{1}{4\sqrt{n}}\right|<\epsilon\quad\text{ if }\quad n>\frac{1}{16\epsilon^2}.$$
Would this be correct?
(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$
We know that $$\sqrt{n^2+n}-n{=(\sqrt{n^2+n}-n){\sqrt{n^2+n}+n\over \sqrt{n^2+n}+n}\\={n\over \sqrt{n^2+n}+n}\\={1\over 1+\sqrt{1+{1\over n}}}}$$therefore$$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|{={1\over 2}-{1\over 1+\sqrt{1+{1\over n}}}<\epsilon}$$which means that $${ 1+\sqrt{1+{1\over n}}}<{1\over {1\over 2}-\epsilon}\\\sqrt{1+{1\over n}}<{{1\over 2}+\epsilon\over {1\over 2}-\epsilon}\\{1+{1\over n}}<{{1\over 4}+\epsilon^2+\epsilon\over {1\over 4}+\epsilon^2-\epsilon}\\{1\over n}<{2\epsilon\over {1\over 4}+\epsilon^2-\epsilon}$$therefore by choosing $$n>{{1\over 4}+\epsilon^2-\epsilon\over 2\epsilon}$$or even $n>{1\over 8\epsilon}$ for small enough $\epsilon$ we obtain $$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|<\epsilon$$