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 $\epsilon &gt;0$.

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 $\epsilon &gt; 0$ there is an $N \in \mathbb{N}$ such that

$n \geq N \implies \mid \frac{n-1}{n-2}- 1 \mid &lt; \epsilon$

Provided $n \geq N$

For all $n&gt;3$, we can remove absolute value.

$\frac{n-1}{n-2} - \frac{n-2}{n-2} = \frac{1}{n-2} \leq \frac{3}{n} &lt;\epsilon$ $\iff \frac{3}{\epsilon}&lt;n$

Thus any integer $N$ greater than $max(3, \frac{3}{\epsilon})$ has the required property

Any feedback would be much appreciated!
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
$$
\left\lvert \frac{n'-1}{n'-2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{3}{N} = \frac{3}{4} \epsilon &lt; \epsilon.
$$
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
$$
\left\lvert \frac{n' - 1}{n' - 2} - 1 \right\rvert = \frac{1}{n' - 2} \leq \frac{1}{ 2 - 2 + \frac{2}{\epsilon}} = \frac{1}{2}\epsilon &lt; \epsilon
$$
Thus the sequence $(x_n) \rightarrow 1$.
$$\lim_{n \to +\infty}\frac{n - 1}{ n- 2}=\lim_{n\to+\infty}\frac{ 1- 1/n}{ 1- 2/n}=1$$