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?