show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$ show that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$
Here is the algebric proof:
We have $a_n=\sqrt{n+1}-\sqrt{n}$, and we want to show  that $\lim a_n=0$.
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{n+1-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}$$
So, when $n\to\infty$, we get $\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$.
Question: I am wondering does episilon-delta method work here as an alternative proof?

The epsilon-delta method requires you to work out how small a $\delta$ is sufficient for a sought $\epsilon$, so you need your calculation anyway. You want to prove$$\forall\epsilon>0\exists\delta>0\left(\forall n>\frac{1}{\delta}\left(\frac{1}{\sqrt{n+1}+\sqrt{n}}<\epsilon\right)\right).$$It suffices to take $\delta=4\epsilon^2$. Or if we use the more typical $\epsilon$-$N$ definition, take $N=\frac{1}{\delta}=\frac{1}{4\epsilon^2}$.
