<s>[INST] Find $\lim _{n\to \infty}\frac{n-1}{n-2}$ I'm taking a university real analysis course and I have been tasked with proving that the sequence $x_n = \frac{n-1}{n-2}$ converges using first principles.
First fix $\epsilon >0$.
Using the algebra of limits we find the limit in question.
$x_n = \frac{n-1}{n-2} = \frac{1-\frac{1}{n}}{1-\frac{2}{n}} \rightarrow x_n = 1 $ as $n\rightarrow\infty$ 
We observe a limit if for any $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that
$n \geq N \implies \mid \frac{n-1}{n-2}- 1 \mid < \epsilon$
Provided $n \geq N$
For all $n>3$, we can remove absolute value.
$\frac{n-1}{n-2} - \frac{n-2}{n-2} = \frac{1}{n-2} \leq \frac{3}{n} <\epsilon$ $\iff \frac{3}{\epsilon}<n$
Thus any integer $N$ greater than $max(3, \frac{3}{\epsilon})$ has the required property
Any feedback would be much appreciated!
 [/INST] Your proof is correct, but I think you should explicitly specify a choice of $N$ as a function of $\epsilon$. Something like this: For any $\epsilon$ and for all $n' \geq N = \max\{\frac{4}{\epsilon}, 3\}$, then
$$
\left\lvert \frac{n'-1}{n'-2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{3}{N} = \frac{3}{4} \epsilon < \epsilon.
$$
Thus the sequence $(x_n) \rightarrow 1$.

Here's another bound that works: For any $\epsilon$ and for all $n' \geq N = \max\{3, \lceil 2 + \frac{2}{\epsilon} \rceil\}$, then
$$
\left\lvert \frac{n' - 1}{n' - 2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{1}{ 2 - 2 + \frac{2}{\epsilon}} = \frac{1}{2}\epsilon < \epsilon
$$
Thus the sequence $(x_n) \rightarrow 1$.
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