Limit proof of a sqrt-heavy expression with binomial formula / sandwich-rule 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?
 A: Why not just multiply/divide by the conjugate, then divide numerator/denominator by $\sqrt{n}$? $$\circ=\frac{2\sqrt{2n}}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}=\frac{2\sqrt{2}}{\sqrt{1+\sqrt{2/n}}+\sqrt{1-\sqrt{2/n}}}\to\frac{2\sqrt{2}}{1+1}=\sqrt{2}.$$
A: First method: use the conjugate quantity.
Multiplying and dividing $b_n$ by $\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}$ yields
$$b_n := \frac{(n+\sqrt{2n})-(n-\sqrt{2n})}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}=\frac{2\sqrt{2n}}{\sqrt{n+\sqrt{2n}}+\sqrt{n-\sqrt{2n}}}.
$$
The denominator is equivalent to $2\sqrt{n}$ hence $b_n\sim2\sqrt{2n}\frac1{2\sqrt{n}}=\sqrt2$.
Second method: use integrals. 
The derivative of $x\mapsto\sqrt{x}$ is the function $\varphi:x\mapsto1/(2\sqrt{x})$, hence
$$
b_n=\int_{n-\sqrt{2n}}^{n+\sqrt{2n}}\varphi(x)\mathrm dx.
$$
The function $\varphi$ is decreasing hence $\varphi(n+\sqrt{2n})\leqslant \varphi(x)\leqslant \varphi(n-\sqrt{2n})$ for every $x$ in the interval of integration. This interval has length $2\sqrt{2n}$ hence, integrating the double inequality for $\varphi(x)$, one gets
$$
2\sqrt{2n}\varphi(n+\sqrt{2n})\leqslant b_n\leqslant 2\sqrt{2n}\varphi(n-\sqrt{2n}).
$$
Since(1) $\varphi(n-\sqrt{2n})\sim \varphi(n+\sqrt{2n})\sim 1/(2\sqrt{n})$, this proves that $b_n\to\sqrt2$.
Third method: use Taylor expansions (see @Phira's answer).
(1) This uses only the fact that $\sqrt{1+z}=1+o(1)$ when $z\to0$, that is, the continuity of the square root function at $1$. To wit, $1/\varphi(n\pm\sqrt{2n})=2\sqrt{n}\sqrt{1+z_n}$ with $z_n=\pm\sqrt{2/n}\to0$ hence $1/\varphi(n\pm\sqrt{2n})\sim2\sqrt{n}$.
A: This can be more or less considered as a variation on Phira's and Didier's answers.
You can write 
$$b_n = \sqrt{n+\sqrt{2n}}-\sqrt{n-\sqrt{2n}}=\sqrt{n}\left(\sqrt{1+\sqrt{\frac2n}}-\sqrt{1-\sqrt{\frac2n}}\right).$$
Then we can use the Mean Value Theorem for the function $f(x)=\sqrt{x}$ on the interval with the endpoints $1-\sqrt{\frac2n}$ and $1+\sqrt{\frac2n}$ which shows that
$$b_n=\sqrt n \cdot 2\sqrt{\frac2n} \cdot \frac1{2\xi_n},$$
where $\xi_n\in(1-\sqrt{\frac2n},1+\sqrt{\frac2n})$. We see that $\xi_n\to 1$ and 
$$\lim\limits_{n\to\infty} b_n = \sqrt2.$$

EDIT:
You description of an alternative method (suggested by your fellow students) is not entirely clear to me, but I understood it as follows:
They noticed that $n\pm\frac{\sqrt2}2=\sqrt{\left(n\pm\frac{\sqrt2}2\right)^2}=\sqrt{n\pm\sqrt{2n}+\frac12}$
In case you succeed in proving
$$\lim\limits_{n\to\infty} (\sqrt{n+\sqrt{2n}}-\sqrt{n+\sqrt{2n}+\frac{\sqrt{2}}2}) = \lim\limits_{n\to\infty} (\sqrt{n-\sqrt{2n}}-\sqrt{n-\sqrt{2n}+\frac{\sqrt{2}}2})=0 \qquad (*)$$ 
then you can use this to show that
$\lim\limits_{n\to\infty} b_n = \lim\limits_{n\to\infty} (\sqrt{n+\sqrt{2n}}-(n+\frac{\sqrt2}2) ) - $ $\lim\limits_{n\to\infty} (\sqrt{n-\sqrt{2n}}-(n-\frac{\sqrt2}2) ) + $ $\lim\limits_{n\to\infty} ((n+\frac{\sqrt2}2) - (n-\frac{\sqrt2}2) ) = \sqrt{2}$
So it remains only question is how to prove $(*)$. If you show it by some of the methods suggested for your original problem then you could have the used this method for the original problem as well. (Und dann ist es also alles umsonnts gewesen.) So I think that if you cannot find some new method to prove $(*)$, this approach did not bring too much. It only added an intermediary step which may or may not help you make things clearer - I believe this depends on your personal taste. Another possible advantage of this approach could be, that if you solved $(*)$ in a similar manner as suggested in the answers, you would probably work with slightly simpler expressions.
A: 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," $\displaystyle\lim_{u\to\infty}\left(\sqrt{u^2+1/2}-u\right)=0$.
A: If you know Taylor expansions, you can use the series for $(1+x)^{\frac12}$ which is probably the binomial formula you are referring to.
We know $(1+x)^{\frac12}= 1+ \frac12 x + O(x^2)$ for small $x$, so $\sqrt{n\pm\sqrt{2n}}= \sqrt n \cdot \sqrt{1\pm\sqrt{\frac2n}}=\sqrt n \cdot (1\pm \frac 12 \sqrt{\frac 2n}+O(\frac 1n))$. The difference of the expression for the two signs gives the desired limits. 
No sandwiches needed for this approach, your fellow students might have referred to the Bernoulli inequality or something similar.
I want to add that I don't see anything "obvious" about the limit.
