Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity Use the ϵ-N definition of limit to prove that lim[(2n+1)/(5n-2)] = 2/5 as n goes to infinity.
The way I do it is
Let ∊ > 0 be given. Notice N ∈ natural number (N) which satisfies {fill this box later}< N.
It follows that if n>=N, then n > {fill this box later}, so for such n, |(2n+1)/(5n-2)-2/5| = |9/(25n-10)| = 9/5|1/(5n-2)|
I am supposed to get to a something that is less than ∊
How to make this to less than ∊?
 A: Try setting your last expression to $\epsilon$ and work backwards. You'll find that the following will work:
Given any $\epsilon > 0$, let $N$ be any natural number that is greater than or equal to the real number:
$$\dfrac{\dfrac{9}{5\epsilon}+2}{5} + 1$$
(by the Archimedean Property, such an $N$ must exist). Now suppose that $n \geq N$ so that:
\begin{align*}
n \geq \dfrac{\dfrac{9}{5\epsilon}+2}{5} + 1
&\implies n > \dfrac{\dfrac{9}{5\epsilon}+2}{5} \\
&\implies 5n - 2 > \dfrac{9}{5\epsilon} \\
&\implies \epsilon > \frac{9}{5} \cdot \frac{1}{5n-2} \\
\end{align*}
Then observe that:
\begin{align*}
\left| \frac{2n + 1}{5n - 2} - \frac{2}{5} \right|
&= \frac{9}{5}\left| \frac{1}{5n - 2} \right| \\
&= \frac{9}{5} \cdot \frac{1}{5n - 2}  \\
&< \epsilon  \\
\end{align*}
as desired.
A: Well, you're correct that you need $$\left|\frac{9}{25n-10}\right|<\epsilon.$$ Note that the given fraction is positive for all $n\in\Bbb N,$ so that we want $$\frac{9}{25n-10}<\epsilon.$$ Try to solve that inequality for $n$, so that you can discover an $N$ that works.
