Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$

Question

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 $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$.

Answer 1

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 $\frac{1}{\sqrt{n^2 + 1} + n} < \frac{1}{2 \sqrt{n}} < \epsilon$

Answer 2

Yes, your reasoning is a correct proof.

Answer 3

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

Answer 4

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 $k$.