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?
 A: 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 $N=\frac{1}{\delta}=\frac{1}{4\epsilon^2}$.
A: 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 $1/4\varepsilon^2$.
A: 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 $n \to \infty$.
A: Just for completeness, the "algebraic proof" in spruce's argument has to be finished off by observing that the functions $n \mapsto \sqrt{n+1}$ and $n \mapsto \sqrt{n}$ both tend to $\infty$ as $n$ tends to $\infty$, hence the same holds for $n \mapsto \sqrt{n+1} + \sqrt{n}$ which implies that $n \mapsto {1 \over \sqrt{n+1} + \sqrt{n}}$ tends to $0$ as $n$ tends to $\infty$. So this proof leaves the $\epsilon$s and $N$s where they belong in low-level lemmas about simple functions and uses intuitive facts about monotonicity to derive the desired result from those lemmas. Unwinding this to get $\epsilon$-$N$ estimates is pointless unless you actually need those estimates.
