Using the $\epsilon$-$N$ to prove: $\lim_{n\to \infty}(\sqrt{1+n}-\sqrt{n})\sqrt{n+\frac12}=\frac 12$

Question

Use the $\epsilon$-$N$ definition of a limit to give a formal proof of: $$\lim_{n\to \infty}(\sqrt{1+n}-\sqrt{n})\sqrt{n+\frac12}=\frac12.$$

I am not sure how to proceed with the $\epsilon$-$N$ definition.

Answer 1

Let $\displaystyle a_n=\frac{\sqrt{n+1/2}}{\sqrt{n+1}+\sqrt{n}}$ be the general term of your sequence. Then:

$$|a_n-1/2|=\frac{|\sqrt{n+1/2}-\sqrt{n+1}+\sqrt{n+1/2}-\sqrt{n}|}{2(\sqrt{n+1}+\sqrt{n})}\leq $$ $$\leq \frac{\sqrt{n+1}-\sqrt{n+1/2}}{2(\sqrt{n+1}+\sqrt{n})}+\frac{\sqrt{n+1/2}-\sqrt{n}}{2(\sqrt{n+1}+\sqrt{n})} =$$ $$=\frac{1}{4(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n+1/2})}+\frac{1}{4(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1/2}+\sqrt{n})} $$ $$\leq \frac{1}{4(\sqrt{n}+\sqrt{n})(\sqrt{n}+\sqrt{n})}+\frac{1}{4(\sqrt{n}+\sqrt{n})(\sqrt{n}+\sqrt{n})}=\frac{1}{8\sqrt{n}}.$$

Let $\epsilon>0$. By the Archimedian property there is $n_0\in \mathbb{N}$ such that $\displaystyle \frac{1}{8\sqrt{n_0}}<\epsilon$, therefore for all $n\geq n_0$: $$|a_n-1/2|\leq \frac{1}{8\sqrt{n}}\leq \frac{1}{8\sqrt{n_0}}<\epsilon.$$

Answer 2

$$\left|\left(\sqrt{1+n}-\sqrt n\right)\sqrt{n+\frac12}-\frac12\right|=\left|\frac{\sqrt{n+\frac12}}{\sqrt{n+1}+\sqrt n}-\frac12\right|=\left|\frac{2\sqrt{n+\frac12}-\sqrt{n+1}-\sqrt n}{2\left(\sqrt{n+1}+\sqrt n\right)}\right|=$$$${}$$

$$=\frac{2\sqrt{n+\frac12}-\sqrt{n+1}-\sqrt n}{2\left(\sqrt{n+1}+\sqrt n\right)}\le\frac{\sqrt{n+\frac12}-\sqrt n}{\sqrt{n+1}+\sqrt n}=\frac{\frac12}{\left(\sqrt{n+1}+\sqrt n\right)\left(\sqrt{n+\frac12}+\sqrt n\right)}\le$$$${}$$

$$\le\frac1{2n}<\epsilon\implies\;\text{enough to take}\;\;n>\frac1{2\epsilon}$$