# Prove $\left\{\frac{n}{2n+3}\right\}$ and $\left\{\frac{n}{2n-3}\right\}$ converge?

Question: prove that the sequences $\left\{\frac{n}{2n+3}\right\}$ and $\left\{\frac{n}{2n-3}\right\}$ converge using the definition.

What I have: I know both of them have limit $1/2$.

The definition says: a sequence $\{x_n\}$ is said to converge to a number $x \in \mathbb R$, if for every $\epsilon >0$, there exists an $M \in \mathbb N$ such that $|x_n-x|<\epsilon$ for all $n\geq M$.

Start: let $\epsilon >0$ such that.... Where do I go from here?

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Then you need to find $M$ such that .... – Arctic Char Nov 5 '13 at 7:32
Let e>0 such that M element N and 1/M+1<e. then for any n>=M ...is this correct thus far? – Cynthia Nov 5 '13 at 7:36
The goal is to get $|\frac{n}{2n+3} -\frac{1}{2}| < \epsilon$. You might play around with this inequality to found how large should $n$ be. – Arctic Char Nov 5 '13 at 7:37

Let $\varepsilon >0$. We want do show the existence of a number $M\in \mathbb{N}$ such that $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$.

$$\left|\frac{n}{2n+3}-\frac{1}{2}\right|=\frac{3}{2(2n+3)}<\varepsilon\iff \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right)<n$$ so if we choose $M> \frac{1}{2}\left( \frac{3}{2\varepsilon}-3\right),$ we get $\left|\frac{n}{2n+3}-\frac{1}{2}\right|< \varepsilon$ if $n\ge M$.

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Would it work the same way for sequence {n/(2*n-3)}? Thank you. – Cynthia Nov 5 '13 at 7:48
@Cynthia Yes, the argument is essentially the same. – Kortlek Nov 5 '13 at 8:03

$\frac{n}{2n+3} = \frac{1}{2+3/n}$ so as n goes to infinity $3/n$ approaches zero. So the limit of the sequence is $\frac{1}{2}$

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I think she might want a $\epsilon$-$N$ approach. Your answer is absolutely correct. – Arctic Char Nov 5 '13 at 7:36
Yea that was my mistake. I just caught that. – RDizzl3 Nov 5 '13 at 7:38