Show (with epsilón - delta) that $\lim_{x -> \infty}(\sqrt{n^2+n}-n)=\frac{1}{2}$

Question

I need show that $\forall \epsilon > 0$ exist $M \in \mathbb{N}$ such that $|\sqrt{n^2+n}-n-\frac{1}{2}| < \epsilon$ for $n \geq M$.

I tried considering $|\frac{(\sqrt{n^2+n}-n)\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n)}-\frac{1}{2}|=|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+n}|<|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}}|$

I need help because now I don't know how abound more... Thanks.

Answer 1

Consider $\sqrt{n^2+n}-n-\frac{1}{2}$. Multiply top and missing bottom by $\sqrt{n^2+n}+n+\frac{1}{2}$. We get $\frac{-1/4}{\sqrt{n^2+n}+n+\frac{1}{2}}$. The absolute value of this is less than $\frac{1}{8n}$. Now finding an appropriate $M$ is straightforward.