# Prove that ${\lim_{n\to\infty}\frac{2n-1}{n+1}}=2.$

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!

• Proof is correct. Apr 12, 2018 at 21:31
• @openspace Thank you. Apr 12, 2018 at 21:37
• I have added (epsilon-delta) tag - since it seems that this is the method of proof you're after. Apr 17, 2018 at 1:49

The argument is good. More directly $$\left|\frac{2n-1}{n+1}-2\right|=\left|\frac{-3}{n+1}\right|$$ so $$\left|\frac{2n-1}{n+1}-2\right|<\varepsilon \quad\text{if and only if}\quad \frac{3}{n+1}<\varepsilon$$ which becomes $$n>\frac{3}{\varepsilon}-1$$ and it's sufficient to find a single $N$ with $N>\frac{3}{\varepsilon}-1$ (which is possible by the Archimedean property) to conclude that, for every $n\ge N$, the required inequality is satisfied.

• Actually this explains my friends result which was $\frac{3-\epsilon}{\epsilon}$. He actually broke down the absolute value in two parts and found two different N. Thank you. Apr 12, 2018 at 22:13
• @YoloSatoshi You're welcome! If $N>\frac{3}{\varepsilon}$, then also $N>\frac{3}{\varepsilon}-1$. You just find a “bigger” $N$ than your friend and it's unimportant: as I say to my students, numbers are for free, so we can keep as large as we prefer. Apr 12, 2018 at 22:15
• Idea however is to find an N that works correct? It does not matter the size, but that it solves the inequality? Apr 12, 2018 at 22:16
• @YoloSatoshi Yes, there's no need to find “the smallest $N$”. Apr 12, 2018 at 22:17

It seems correct, as an alternative we could note that

$$\lim_{n\to\infty}\frac{2n-1}{n+1}=\lim_{n\to\infty}\frac{2n+2-3}{n+1}=\lim_{n\to\infty}2-\frac{3}{n+1}=2$$

and prove that

$$\lim_{n\to\infty} \frac{3}{n+1}=0\iff \lim_{n\to\infty} \frac{1}{n+1}=0\iff \lim_{n\to\infty} \frac{1}{n}=0$$

Or one could write

$\dfrac{2n - 1}{n + 1} = \dfrac{2 - n^{-1}}{1 + n^{-1}} \to 2 \; \text{as} \; n \to \infty. \tag 1$