Prove that $\displaystyle{\lim_{n\to\infty}\frac{2n-1}{n+1}}=2.$
Proof: Let $\epsilon>0$. We want to show that $\exists N\in\mathbb{N}$ such that. $$ n\geq N\Rightarrow \left|\frac{2n-1}{n+1}-2\right|<\epsilon.$$
Using the archimedean property, we can find a positive integer $N$ such that $N>\frac{3}{\epsilon}$ . We will next show that this $N$ works. Let $n\geq N$. Then, $$\left|\frac{2n-1}{n+1}-2\right|=\left|\frac{2n-1}{n+1}-\frac{2(n+1)}{n+1}\right|=\left|\frac{2n-1-2n-2}{n+1}\right|=\left|\frac{-3}{n+1}\right|=\frac{3}{n+1}<\frac{3}{n}\leq\frac{3}{N}<\epsilon $$
Hence, by definition it follows that $\displaystyle{\lim_{n\to\infty}\frac{2n-1}{n+1}}=2.\blacksquare$
My friend in grad school whom I respect greatly did it completely differently and found a different result so I wanted to make sure. Thank you!