What's the limit of the sequence $a_{n} = \frac{\lfloor\sqrt{2n}\rfloor}{\lfloor\sqrt{n}\rfloor}$ ?
So i know that the answer is $\sqrt{2}$ but how do i prove it? I tried using the sandwich theorem but failed to find a bigger sequence that approached $\sqrt{2}$, I also tried proving it from definition but got nowhere. I can't really think of anything to do about the floor function since it disrupts continuity. I can prove that the subsequence $a_{n^2}$ converges to $\sqrt{2}$ but i don't think that helps much.
Hint: using the fact that, for all $x$, we have $\lfloor x\in (x-1, x]$, we have $$a_n = \frac{\lfloor\sqrt{2n}\rfloor}{\lfloor n\rfloor} \leq \frac{\sqrt{2n} + 1}{\sqrt n - 1}$$
and
$$a_n = \frac{\lfloor\sqrt{2n}\rfloor}{\lfloor n\rfloor} \geq \frac{\sqrt{2n} - 1}{\sqrt n + 1}$$
which should make the situation much easier to handle.