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?

Observe that $\sqrt{n+1} > \sqrt{n}$, so, $\sqrt{n+1} + \sqrt{n} > 2\sqrt{n}$ and then
$$\frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{2\sqrt{n}} \ .$$
Now, let $\varepsilon>0$. Since $4\varepsilon^2$ is a positive real number, by the Archimedian property of $\mathbb{R}$ there exists $N \in \mathbb{N}$ such that
$$0< \frac{1}{N} < 4\varepsilon^2.$$
It follows that for $n \geq N$, $2\sqrt{n} \geq 2\sqrt{N}$ and then
$$a_n = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \dfrac{1}{2\sqrt{n}} \leq \dfrac{1}{2\sqrt{N}} < \varepsilon.$$
Since $\varepsilon$ was arbitrary, we have shown that $a_n$ can make arbitrarily small for sufficiently large $n$. Hence, $a_n \to 0$ as $n \to \infty$.
