# Advantages of the sequence definition of limits

I will be teaching an introductory analysis course in the coming semester. In it the students will learn about limits of real sequences, and then will learn about limits of functions in terms of sequences.

More precisely, we will say that $\lim_{x\to a+}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n>a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Likewise, we will say that $\lim_{x\to a-}f(x) = L$ if whenever $(x_n)$ converges to $a$ with $x_n< a$ for all $n$, we have $\lim_{n\to \infty} f(x_n) = L$. Then we say that $\lim_{x\to a} f(x) = L$ if both $\lim_{x\to a+}f(x) = L$ and $\lim_{x\to a-}f(x) = L$.

The students will have already seen the $\varepsilon$-$\delta$ definitions of limits of functions in their calculus course. The question then is, how to properly motivate this second (equivalent) definition of limits of functions?

Are there any arguments which become significantly easier when using the sequence definition of limits of functions in place of the $\varepsilon$-$\delta$ definition? (These should be elementary enough to be understood by first year Mathematics undergraduates.)

For instance, I suppose that once one has the Algebra of Limits for sequences, one gets the Algebra of Limits for functions for free. But I'm not convinced that much is to be gained from doing things this way around.

Edit: Thanks for the answers and comments so far. It seems many people are in favour of teaching the sequence definition of limits alongside the $\varepsilon$-$\delta$ definition. I agree that it should be useful to be aware of both definitions. To be certain of this, however, I would still like to see an example of a proof which is simpler when using the sequence definition.

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A common approach, as pointed out in an answer below, is to teach limits of sequences first and then to define limits of functions using sequences, in an attempt to use as few $\epsilon - \delta$ as possible (basically, just enough to prove that $1/n$ converges to $0$ and to prove what you call the "algebra of limits" -- after that, you can get away without ever seeing an $\epsilon$ again in the first year, if you so wish). In your case though, if the students have seen epislons and deltas and do not hate them (as many students do, I have found), there is little motivation to (to be continued) – Pierre Aug 30 '12 at 12:00
little motivation to give the alternative definition. Nevertheless, you could argue that it gives a "dynamical system" feel to the concept of limit, and that can be pretty visual and intuitive with some students. – Pierre Aug 30 '12 at 12:02
There was an MSE question on this pedagogical point, but I have mislaid the bookmark – Yemon Choi Aug 30 '12 at 23:36
Yemon might have math.stackexchange.com/q/113698 or math.stackexchange.com/q/186274 in mind, both containing some relevant remarks. – Theo Buehler Aug 30 '12 at 23:47

## 5 Answers

Interestingly enough, as far as I remember, we did limits of sequences first (without functions) and then limits of functions (with $\varepsilon-\delta$) and then, as a remark, the connection mentioned above behind the very same Iron Curtain. Actually, in the end of our "limits course", our professor either did limits over nets, or stopped one step short of it: he certainly said all necessary words and made it clear that to talk about of a limit of a mapping, you just need some set of "catchers" in the range space and some set of "tails" in the argument space. That was a bit tough in the beginning but paid off nicely when doing Riemann integration where the tails are either partitions of small mesh or partitions subordinated to a fixed partition. I still find this abstract view rather enlightening; much more enlightening that the lemma in question, which, IMHO, only makes the concept more confusing (though is quite useful as a technical tool). The main reason for this opinion is that this abstract view is unifying: all notions of limit that the students will ever meet fall under this idea, only the choices of catchers and tails vary and only one magic phrase is ever needed: "For every catcher, there is a tail whose image is contained in the catcher". The lemma you mentioned is separating: if used as a definition rather than a remark, it creates an impression that there are many ad hoc concepts of limits that all have to be understood and memorized separately, creating quite a mess in the student's head. The $\varepsilon-\delta$ definition is already hard because it mixes the limit concept and the technical descriptions of the catchers and the tails, i.e., 3 things that can be easily separated and on which you can train the students one by one if you start with the abstract view. To be honest, I haven't tried it myself in the USA yet but I certainly will when teaching freshman analysis (so far it was either business calculus, where the game is never worth the candles, or advanced courses where the concept of limit was assumed to be well-known already).

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+1 for "the very same Iron Curtain". But I don't get the "catchers" and "tails" business. What's the mental image here? – Tom Leinster Aug 30 '12 at 11:55
@fedja: Thanks. By "the lemma in question", are you referring to the fact that the sequence and $\varepsilon$-$\delta$ definitions of limit of a function coincide? If so, could you expand on your final sentence (in particular the part in parentheses)? – Mark Grant Aug 30 '12 at 11:56
@Tom I'll answer first in the formal language: We have $f:X\to Y$. To set up a limit, we need $T\subset S(X)$ (tails) and $C\subset S(Y)$ (catchers) ($S(Z)$ is the set of all subsets of $Z$). Then we just say that "$f$ has a limit" if for every $B\in C$ there exists $A\in T$ with $f(A)\subset B$ (whenever the argument is restricted to some tail, the value is caught in the given catcher). That's the underlying idea. – fedja Aug 30 '12 at 12:17
The rest are just choices: for instance, if $C$ is the set of all open sets containing a set $K$ and $T$ is the set of truncated angles around a given ray in $R^n$, we say that $f$ is attracted to $K$ when the argument escapes to infinity along a given direction. The beauty, as I said, is in one concept serving all cases. To have it meaningful, a couple of simple axioms are needed like "the intersection of two tails contains some tail" and "all tails are non-empty", but after that you are free to design anything you want. – fedja Aug 30 '12 at 12:21

I can only speak of my experience. I was first taught convergence of sequences and then later on, the definition of limits with $\varepsilon-\delta$ and its equivalent form involving sequences. I still find this most intuitive, but then this opinion is clearly heavily influenced by my upbringing.

For disclosure, I received my math education behind the Iron Curtain, and this model was employed by most (now former) communist countries. During that time curricula in most of those countries were devised by influential mathematicians who could also communicate math very well. (For example, in USSR Kolmogorov was deeply involved in shaping math education. He even wrote some high-school textbooks that were widely used.)

What I am attempting to communicate here is that the sequences-first system was adopted by informed mathematicians who cared about math education, it was tested on a large scale (tens if not hundreds of millions of students) for a long time (several decades). Arguably this system has produced good results.

Terry Tao's textbook on analysis (which I like very much for several reasons) also relies on a sequences-first approach.

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@Liviu: Thank you for your answer. Certainly I am not knocking this order of doing things from a pedagogical viewpoint. My only concern is that when one gives a definition of a concept, then later gives a second equivalent definition of the same concept, one should motivate the second definition by showing that it makes life easier than the first definition in some cases. Otherwise, the lazier students will ask why they need to learn two definitions of the same concept (and I won't have a good answer). – Mark Grant Aug 30 '12 at 11:22
Of course, when you prove the equivalence of the epsilon-delta definition and the sequential definition, you are exposing the students to the axiom of choice at a rather early stage in their education. – bof Dec 21 '15 at 11:59
@bof You can avoid the axiom of choice. The hard part, the sequence def implies epsilon-delta definition can be argued by contradiction. Conclude by choosing delta of the form $1/n$. – Liviu Nicolaescu Dec 21 '15 at 12:05
No, I don't believe you can prove that implication without the axiom of choice. – bof Dec 21 '15 at 13:04
At p. 98-99 in these notes www3.nd.edu/~lnicolae/Hon_Calc_Lectures.pdf there is a proof that does not use the axiom of choice. – Liviu Nicolaescu Dec 21 '15 at 14:11

In teaching a second year UK course in analysis some years ago I was surprised to find out how pleasant were the proofs of the some of the basic results using sequential methods, for example that a continuous function on a closed bounded subset of $\mathbb R^n$ to $\mathbb R$ is bounded. This inspired me to work out a paper on the notion of a $1$-point sequential compactification: add another point and let the sequences in $X$ with no convergent subsequence converge to the extra point! It got published too in the JLMS.

However for continuity of a function I do like to rely on the neighbourhood definition since a neighbourhood is a geometric object one can draw, whereas the $\epsilon - \delta$ are only measurements of the sizes of neighbourhoods.

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"Are there any arguments which become significantly easier...". There is an obvious choice: Disprove that a certain function has a limit at a certain point. Showing that with epsilon-delta would be a mission impossible for most students. Using the sequence definition makes it much easier and more appealing.

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This is a good response to Christian Blatter's answer. It may be harder to prove continuity, but by the same token, it is much easier to disprove it. – Ryan Reich May 5 '13 at 20:41

In my view there are no advantages in defining function limits via sequences, and this practice should be abolished. Using this definition you would have to test $\aleph_1^{\aleph_0}$ or so sequences to prove a single instance of $\lim_{x\to a} f(x)=\alpha$. Why should one bring all these sequences into the picture?

The idea of "limit of $f(x)$ when $x\to a$" is the answer to the following question: What is the "natural" value of $f$ at the special, maybe "ideal", point $a\$? Well, it's the value that would make $f$ continuous there.

This brings me to the main point: The primary and sufficiently intuitive notion is the notion of continuity. Unfortunately the simple concept of Lipschitz continuity does not cover all cases we'd like to handle, e.g. $\sqrt{|x|}$ at $x=0$. Therefore we have to dig deeper and come up with $\epsilon$ and $\delta$, and on, and on.

Sequences, on the other hand, are a fundamental tool to construct new objects, like $e$, or $\sqrt{2}$. Of course in passing we would then prove that $\lim_{x\to a}f(x)=\alpha$ iff for all sequences $\ldots$

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Sorry, but the argument on the cardinality of the set of sequences one needs to test makes no sense to me. Why don't you have a problem with having to test for each of the $2^{\aleph_0}$ (assuming this is the cardinailty of the continum) many values of $\varepsilon$ that something holds for the typically $2^{\aleph_0}$ many $x$ in some $\delta$-neighborhood? – user9072 Aug 30 '12 at 20:43
The above being said, while I am not fully decided on the general matter, I am aware of the fact that some students have difficulty with the 'for each sequence' and so on, for instance just showing something for one specific choice of sequence and thinking they are done. It is just this cardinality argument that I cannot follow, not the general point of view. – user9072 Aug 30 '12 at 20:59
Not to insists too much on what is essentially tangential to your argument (in my mind), yet since you changed the way you write the cardinality: even if it now says $\aleph_1^{\aleph_0}$ , this is still $2^{\aleph_0}$, the cardinality of the continuum (except perhaps in some more exotic set theories), and you won't get the thing 'cheaper' than with considering continuum many things somewhere. I think there really is no argument to be made based on cardinalities. To insist on brining this aspect in, in my opinion takes away from an (otherwise) very reasonable argument. – user9072 Aug 31 '12 at 12:27
@quid: Of course the thing with the cardinality was meant to be a joke. The essential point is that bringing in sequences replaces a simple limit by an infinity of composite ("verschachtelte") limits, over which reigns an additional $\forall$. – Christian Blatter Aug 31 '12 at 13:00
Oh, sorry, for missing the tongue-in-cheek nature of that part of the argument. This makes more sense then. By the way, I like your books on the subject. – user9072 Aug 31 '12 at 13:20