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

Question

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!

Answer 1

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.

Answer 2

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

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

Answer 3

Or one could write

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