TITLE: Limit proof of a sqrt-heavy expression with binomial formula / sandwich-rule
QUESTION [6 upvotes]: Let be
$$b_n := \sqrt{n+\sqrt{2n}}-\sqrt{n-\sqrt{2n}}, n\in\mathbb{N}$$
a sequence. I am to determine $\lim\limits_{n\to\infty}b_n$ which is obviously $\sqrt{2}$.
My first step was to transform the sequence to $-\sqrt{(-\sqrt{2}+\sqrt{n})\sqrt{n}}+\sqrt{(\sqrt{2}+\sqrt{n})\sqrt{n}}$ however this didn't helped me out. Some fellow students used some funky binomial formula to bound the limit to a specific range via the "sandwich rule" (I didn't found a suitable translation from german to english) which is defined by:
Definition. Let $a_n\leq a_n'\leq a_n''$ for nearly all $n\in\mathbb{N}$. Then one has
$$\lim\limits_{n\to\infty}a_n = \lim\limits_{n\to\infty}a_n''=a\Rightarrow\lim\limits_{n\to\infty}a_n'=a.$$
I do understand the basic idea to find an expression similiar to $\sqrt{n\pm\sqrt{2n}}$ to be able to bound the limit to a range however it doesn't seem obvious to me what I can do here. My second idea was to show that $b_n$ is bounded by $[0;2)$ and then to conclude, that $\sqrt{2}$ is the proper limit.
Is anyone able to explain the idea with the binomial formula and the sandwich-rule?
EDIT
My fellow students tried to prove $\lim\limits_{n\to\infty}\sqrt{n\pm\sqrt{2}}=\lim\limits_{n\to\infty}\sqrt{n\pm\sqrt{2}+0.5}=\lim\limits_{n\to\infty}\sqrt{(n\pm\sqrt{2}/2)^2}$ however I do not understand how to develop this idea. Their next step was to show $\lim\limits_{n\to\infty}b_n=\sqrt{2}$ via $\lim\limits_{n\to\infty}\left(\sqrt{n\pm\sqrt{2}}-(n\pm\sqrt{2}/2)\right)=0$. Can anyone tell me whether this is correct and explain what is the clue behind this?

REPLY [2 votes]: The following is in response to the edited question. 
It may be interesting to complete the square, in order to bring out the symmetry.  Note that
$$n+\sqrt{2n}=\left(\sqrt{n}+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}
\qquad\text{and}\qquad n-\sqrt{2n}=\left(\sqrt{n}-\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}.$$
Maybe completing the square qualifies as a "funky binomial formula."
To save some typing, we write $x$ for $\sqrt{n}+\frac{1}{\sqrt{2}}$ and $y$ for
$\sqrt{n}-\frac{1}{\sqrt{2}}$.  Note that $x-y=\sqrt{2}$, and therefore
$$\sqrt{x^2+1/2}-\sqrt{y^2+1/2}=\left(\sqrt{x^2+1/2}-x\right)-\left(\sqrt{y^2+1/2}-y\right)+\sqrt{2}.$$
As $n$ grows without bound, so do $x$ and $y$. Thus it will be enough to show that $\displaystyle\lim_{u\to\infty}\left(\sqrt{u^2+1/2}-u\right)=0$.  
This can be proved using any of the several ideas described in the other answers. For example, if $u$ is positive, then
$$u^2+\frac{1}{2}<\left(u+\frac{1}{4u}\right)^2,$$
(just expand the right-hand side) and therefore 
$$0<\sqrt{u^2+1/2}-u <u+\frac{1}{4u}-u=\frac{1}{4u}.$$
Thus, by "squeezing," $\displaystyle\lim_{u\to\infty}\left(\sqrt{u^2+1/2}-u\right)=0$.