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 $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$.
 A: Yes, your reasoning is a correct proof.
A: 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 $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 \sqrt{n}} < \epsilon$
A: The typical technique here is to get this into a fraction rather than a subtraction.  Here, multiply by the 'conjugate' of the top,  so multiply it by $\frac {\sqrt {n^2+1} +n} {\sqrt {n^2+1} +n}$
Then you get $\frac 1 {\sqrt {n^2+1} +n}$.   Now you have a limit that is of the form $\frac 1 \infty$, so it goes to 0
A: Another way:
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
$.
For $0 < x $,
$(1+x)^2
=1+2x+x^2
> 1+2x
$,
so
$\sqrt{1+2x} < 1+x$
or
$\sqrt{1+x} < 1+x/2$.
(You can directly prove this
by squaring
both sides.)
Putting
$x = 1/n^2$,
$\sqrt{1+1/n^2}
< 1+1/(2n^2)
$.
Therefore,
$\sqrt{n^2+1}
=n \sqrt{1+1/n^2}
< n(1+1/(2n^2))
= n+1/(2n)
$,
so
$\sqrt{n^2+1}-n
< (n+1/(2n))-n
=1/(2n)
\to 0$
as $n \to \infty
$.
Note that can 
be used to show that
$\sqrt{n^2+k}-n
\to 0$
for any fixed $k$.
