I'm doing a university real analysis course and I'm practising proving the limits of sequences. I've been tasked prove $x_n = \frac{n-1}{n^2+2}$ converges.
Using algebra of limits we deduce.
$$x_n =\frac{n-1}{n^2+2} = \frac{\frac{1}{n}-\frac{1}{n^2}}{1+\frac{2}{n}} \implies x_n=0$ as $n \rightarrow \infty.$$
A limit exists if for any $\epsilon>0$ there is a $N \in \mathbb{N}$
$$n \geq N \implies |\frac{n-1}{n^2-2}-0| < \epsilon.$$
For all $n\geq 1$ we can remove the absolute value.
$$\frac{n-1}{n^2-2} \leq \frac{n-1}{n} = 1-\frac{1}{n} < \epsilon \iff n<\frac{1}{1-\epsilon}.$$
Thus any $N$ greater than $\max\{1,\frac{1}{1-\epsilon}\}$ has the required property.
Any feedback would be much appreciated!
Here is a proof only using the definition.
Let $\varepsilon>0$ be arbitrary. Pick an $N$ such that $N>1/\varepsilon$. For any $n\geq N$ $$\left\lvert\frac{n-1}{n^2+2}\right\rvert\leq\frac{n}{n^2+2}\leq\frac{n}{n^2}=\frac{1}{n}\leq\frac{1}{N}<\varepsilon.$$