<s>[INST] Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$ I'm trying to prove $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$. Is the following a correct proof?
For all $n$ we have $0 \leq \left|\sqrt{n^2 + 1} - n\right| \leq \left|\sqrt{n^2+1} - 1 \right|$. For any $\epsilon > 0$ take $ N = \sqrt{(\epsilon+1)^2-1}$. Then for all $n > N$ we have $\left|\sqrt{n^2+1} - 1 \right| < \epsilon$ so $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - 1] = 0$ and by the squeeze theorem $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$.
 [/INST] No, it isn't. First of all think: does $\lim \limits_{n \to \infty} \sqrt{n^2 + 1} - 1 = 0$?
Your mistake is that assuming that when we take $n > N$, that we will have $\sqrt{n^2 + 1} - 1 < \epsilon$, but actually when we take $n > N$ we will have $\sqrt{n^2 + 1} - 1 > \epsilon$.
Here would be a correct proof:
Multiply by $\frac{\sqrt{n^2 + 1} + n}{\sqrt{n^2 + 1} + n}$ to get $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 n}$. Thus if we pick $N = (2 \epsilon)^{-1}$, then for all $n>N$ we will have $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 \sqrt{n}} < \epsilon$
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