Understanding the ##ε## as used in limits of sequences

[chwala]

TL;DR Summary

Showing that for any given $ε>0$ (no matter how small), there exists a number
$N$( depending on $ε$) s.t $|U_n - l| <ε$

I will create my own example on this- Phew atleast this concepts are becoming clearer ; your indulgence is welcome.
Let me have a sequence given as,

$Un = \dfrac {7n-1}{9n+2}$

$Lim_{n→∞} \left[\dfrac {7n-1}{9n+2}\right] = \dfrac {7}{9}$

Now,

$\left[ \dfrac {7n-1}{9n+2} - \dfrac {7}{9} \right] = \left[ \dfrac {-23}{9(9n+2)} \right]$

when,

$\left[ \dfrac {23}{9(9n+2)} \right] <ε$

or

$\left[ \dfrac {9(9n+2)}{23} \right] >\left[\dfrac{1}{ε}\right]$

$9n+2 >\dfrac{23}{9ε}$

$9n > \dfrac{23}{9ε} -2$

$n > \dfrac{1}{9} \left[ \dfrac{23}{9ε} -2 \right]$

choosing $N= \dfrac{1}{9} \left[ \dfrac{23}{9ε} -2 \right]$ where say for e.g Let $ε = 0.01$

$N= \dfrac{1}{9} [ 255.55555-2]=253.5555$

This means that, all terms beyond $253$ differ from $\dfrac {7}{9}$ by an absolute value less than $0.01$.

Let us check,

If $N=300, u_n =\left|\dfrac{2099}{2702} - \dfrac {7}{9}\right|=|0.77683-0.7777|=0.00087<0.01$

Implying that if i pick a sequence less than $253$ then the proof will not hold. This is the time that i am getting to understand some analysis ...particularly of this epsilon thing!

Cheers. Any insight welcome.

Maybe the question that i may need to ask is how small can $ε$ be and how big can it be? I could say less than $1$ in order for the limit to exist.

 

[fresh_42]

Maybe the question that i may need to ask is how small can $ε$ be and how big can it be? I could say less than $1$ in order for the limit to exist.

$\varepsilon$ can be any positive number no matter how big or how small. You can restrict $N(\varepsilon)$ but not $\varepsilon.$ Proofs often have a setting of $N(\varepsilon)$ as the maximum of something, e.g. $N(\varepsilon)=\max\left\{2,\dfrac{1}{9}\left|\dfrac{23}{9\varepsilon}\right|\right\}.$ You do not need the tightest number like the $-2/9$ in your setting. As long as you make $N(\varepsilon)$ bigger, there won't be a problem. I often add $1$ simply because I do not want to bother in which direction the rounding takes place or whether something is equal or strictly less. Be generous with $N(\varepsilon)$ but do not touch $\varepsilon.$

Proofs are usually written twice: one direction to find $N(\varepsilon)$ as you did, but then the other way around, which is the important one! You left that one out, which is a mistake. You have to show $\left|\dfrac{7n-1}{9n+2}-\dfrac{7}{9}\right|< \ldots\ \, < \, \ldots \, < \, \varepsilon$ by using your setting for $N(\varepsilon)$ only.

Let us check,

If ...

is what you really have to do. And not for a specific example, but rather for any choice of $\varepsilon>0.$ The first step of finding $N(\varepsilon)$ is your private pleasure. The proof has to be written "backward". Here is a similar example: https://www.physicsforums.com/insights/epsilontic-limits-and-continuity/

 

[chwala]

Appreciated on your article @fresh_42