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?

You want to prove that for all $\varepsilon > 0$ there is $N > 0$ such that 
$\sqrt{n+1}-\sqrt{n}<\varepsilon$ when $n>N$. Your algebraic manipulation shows that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
so you would like to solve the inequality $\frac{1}{\sqrt{n+1}+\sqrt{n}} < \varepsilon$.
To find $N$ it is enough to note that $\sqrt{n+1}+\sqrt{n} > 2\sqrt{n} > \varepsilon^{-1}$, and the last gives $n > 1/4\varepsilon^2$. Set $N$ to be the least integer greater than $1/4\varepsilon^2$.
