Can you help me understand this definition for the limit of a sequence? I'm reading the textbook "Calculus - Early Transcendentals" by Jon Rogawski for my Calculus III university course.
I'm trying for the life of me to understand the wording of this definition, and I wonder if it can be said in simpler terms to get the basic point across.

A sequence $a_n$ converges to a limit $L$, and we write
$$\lim_{n\to\infty} a_n=L$$
if, for every $\epsilon > 0$, there is a number $M$ such that $|a_n - L| < \epsilon$ for all $n > M$. If no limit exists, we say that ${a_n}$ diverges.

It looks like a very straightforward rule, but I just can't make sense of all the variables in the definition.

Let’s call $\{a_n,a_{n+1},a_{n+2},\dots\}$ the $n$-tail of the sequence. 
Now suppose that I give you a target around the number $L$: I pick some positive leeway $\epsilon$  want you to hit the interval $(L-\epsilon,L+\epsilon)$. We’ll say that the sequence hits that target if some tail of the sequence lies entirely inside the interval. 

For instance, if $a_n=\frac1{2^n}$, the $4$-tail of the sequence hits the target $\left(-\frac1{10},\frac1{10}\right)$ with leeway $frac1{10}$ around $0$: the $4$-tail is $$\left\{\frac1{2^4},\frac1{2^5},\frac1{2^6},\dots\right\}=\left\{\frac1{16},\frac1{32},\frac1{64},\dots\right\}\;,$$ and all of these fractions are between $-\frac1{10}$ and $\frac1{10}$.

It’s not hard to see that no matter how small a leeway $\epsilon$ I choose, some tail of that sequence hits the target $(-\epsilon,\epsilon)$: I just have to find an $n$ large enough so that $\frac1{2^n}<\epsilon$, and then the $n$-tail will hit the target.
Of course, in my example the $4$-tail of the sequence also hits the target $\left(0,\frac18\right)$ with leeway $\frac1{16}$ around $\frac1{16}$. However, there are smaller targets around $\frac1{16}$ that aren’t hit by any tail of the sequence. For instance, no tail hits the target $\left(\frac1{16}-\frac1{32},\frac1{16}+\frac1{32}\right)=\left(\frac1{32},\frac3{32}\right)$: no matter how big $n$ is, $$\frac1{2^{n+6}}\le\frac1{2^6}=\frac1{64}\;,$$ so $\frac1{2^{n+6}}$ is in the $n$-tail but not in the target.
When we say that $\lim\limits_{n\to\infty}a_n=L$, we’re saying that no matter how small you set the leeway $\epsilon$ around $L$, the centre of the target, some tail of the sequence hits that tiny target. Thus, $\lim\limits_{n\to\infty}\frac1{2^n}=0$, and $\lim\limits_{n\to\infty}\frac1{2^n}\ne\frac1{16}$: no matter who tiny a target centred on $0$ you set, there is a tail of the sequence that hits it, but I just showed a target around $\frac1{16}$ that isn’t hit by any tail of the sequence.
One way to sum this up: $\lim\limits_{n\to\infty}a_n=L$ means that no matter how small an open interval you choose around the number $L$, there is some tail of the sequence that lies entirely inside that interval. You may have to ignore a huge number of terms of the sequence before that tail, but there is a tail small enough to fit.
