Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

Basically, what this definition says is that no matter how large or small you pick $\epsilon$, there is some index $M$ such that $a_{M+1}, a_{M+2}, \ldots$ are all within $\epsilon$ of a limit $L$.

So, you have an infinite sequence, and it doesn't matter how small you pick $\epsilon$, you can always find an index $M$ such that every term beyond $a_M$ in the sequence is less than $\epsilon$ units away from some number $L$.

Example: Let's say you pick $\epsilon = 0.001$. Well, then I can pick, say, $M = 50$, and every $a_n$ where $n > 50$ is within $\epsilon$ units of $L$. So you pick $\epsilon = 0.0000001$. Then I might have to pick $M = 43578$. No matter how small an $\epsilon$ you pick, I always win the game.

In practice, we don't often compute actual values of $\epsilon$ and $M$ -- most of the time we can't. But we have other tools to show that it is possible to do so (even if we don't know what the values are).

share|cite|improve this answer
I think I understand this now. The professor had worded it similar, but I couldn't see the link in his wording to the formal definition. He had stated "A sequence An has the limit L and we write ... if the terms An get as close to L as desired by taking n sufficiently large." – agent154 Sep 8 '12 at 1:12

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.

share|cite|improve this answer

Try to prove this first:

Given $a,b$, then $a=b$ if and only if for very $\epsilon >0$, $|a-b|<\epsilon$.

You have probably heard that we might refer to the distance from a number $a$ to another number $b$ by $d(a,b)=|a-b|=|b-a|$. The definition then says that a sequence of numbers $a_n$ has a limit $L$ if, the distance $|a_n-L|$ can be made as small as we wish "for every $\epsilon >0$" if we seek further in to the sequence "for all $n>M$".

Look at the image. The points are the corresponding values of $a_k$ (don't worry about scaling). Since $|a_n-\mathscr L|<\epsilon$ is the equivalent to $-\epsilon<a_n-\mathscr L<\epsilon$, the idea we want to capture is that we can box the terms of the sequence inside thethe strip of width $2\epsilon$ centered at $\mathscr L$ if we take $n$ large enough. Not that since $\epsilon'<\epsilon$ we need to go further on the sequence, and take an $M'$ greater that the original $M$ that worked before.

enter image description here

Maybe this helps. Consider the sequence $$a_n=\frac{1}{n}$$

If I give you $\epsilon =0.000001$, what $M$ you should choose so that $$\left|\frac{1}{n}-0\right|<0.000001?$$

Do you agree that, in this case $$\lim_{n\to \infty}\frac{1}{n}=0?$$

share|cite|improve this answer
Yes - I do understand that the limit of 1/n = 0 in this case. – agent154 Sep 8 '12 at 1:11
@agent154 Well, what about $$\frac{n}{n+1}$$ – Pedro Tamaroff Sep 8 '12 at 1:33
Same thing - (n/n)/((n/n)+(1/n)) becomes 1/1+0 = 1. I do understand how simple limits work; I was just trying to understand what was meant by that verbose rule. – agent154 Sep 8 '12 at 14:29
@agent154 It is not a verbose "rule", it is a formal definition, and you should learn it, understand it, and learn to apply it. It will serve you a lot. What I wanted you to do in the last one is $$\frac{n}{{n + 1}} = 1 - \frac{1}{{n + 1}}$$ and apply your last result. You're not applying the definition in that algebraic manipulation (unless you have proven some algebraic properties about limits of sequences already). – Pedro Tamaroff Sep 8 '12 at 14:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.