Proving the limit of a sequence converges using the definition of a limit So I have $\cdot \lim \limits_{n \to \infty} \sqrt{n^2+1} -n$
I understand that the limit is 0 because,
$$\cdot \lim \limits_{n \to \infty} \sqrt{n^2+1} -n = \left(\sqrt{n^2+1}-n \right)\times\frac{\sqrt{n^2+1}+n}{\sqrt{n^2+1}+n} = \frac{1}{\sqrt{n^2+1}+n}\le \frac{1}{2n}\le\frac{1}{n} = 0$$
But what can I set my function $N(\epsilon)$ to in order for this to be true?
 A: You have already done all the work. Let $\epsilon\gt 0$. You have shown that if $n$ is a positive integer, then $0\lt \sqrt{n^2+1}-n\lt \frac{1}{n}$. We want $|\sqrt{n^2+1}-n|\lt \epsilon$. This will hold if $\frac{1}{n}\le \epsilon$, or equivalently if $n\ge 1/\epsilon$.
Let $N_\epsilon=\lceil 1/\epsilon\rceil$. Here $\lceil x\rceil$ is the ceiling function, the smallest integer which is $\ge x$.
Remark: You had actually proved the stronger result that the absolute value is $\lt \frac{1}{2n}$. So we can actually choose $N=\lceil 1/(2\epsilon)\rceil$.  But there is no particular sense in attempting to choose "best possible" $N$. Sharp estimates can wait until you are doing numerical analysis.
A: Let $x_n=\sqrt{n^2+1}-n$. Write
\begin{align}
x_n &= \exp^{\log x_n}\\ &= \exp^{\log(n^2+1)^\frac12-\log n}\\ &= \exp^{\frac12\log\left(\frac{n^2+1}{n^2}\right)}\\
&=\exp^{\frac12\log\left(1+\frac{1}{n^2}\right)}
\end{align}
By continuity of $\exp$ and $\log$, it follows that
$$\lim_{n\to\infty}x_n =\exp^{\frac12\lim_{n\to\infty}\log\left(1+\frac1{n^2}\right)} = \lim_{t\to -\infty}e^t = 0. $$
