Is the $\epsilon-\delta$ definition of a limit useful for either calculating or proving limits?

Question

I found this question and its subsequent answer extremely helpful in its example of the use of the $\epsilon-\delta$ definition to demonstrate the value of a limit. I also notice that the example itself seems to me to be quite easy to demonstrate without recourse to the $\epsilon-\delta$ definition.

Having said that, is there value in this definition beyond mathematical rigour? Are there limit problems (in single variable calculus) that require the $\epsilon-\delta$ definition in order to either find a limit, demonstrate that one exists, or to prove a result?

Answer 1

Consider the function $f: \Bbb R \to \Bbb R$ defined by $f(x) = \dfrac{1}{q}$, if $x = \dfrac{p}{q} \in \Bbb Q$ with $\gcd(p,q) = 1$, and $f(x) = 0$, for all $x \not\in \Bbb Q$.

It is hard to imagine how one would prove $\lim\limits_{x \to 0} f(x) = 0$ without recourse to the epsilon-delta definition.

That said, using the definition is cumbersome. Building up a repertoire of continuous functions makes things much easier, because we can evaluate limits by simple substitution, in many cases. In a sense, this "shifts the difficulty" into establishing our functions are continuous, but many commonly used functions can be "broken up" into simpler components via sums, multiplying by a constant, multiplying functions together, or through composition, enabling us to prove continuity for a few simple cases, and continuing from there.

Indeterminate forms like $\dfrac{0}{0}$, or $\dfrac{\infty}{\infty}$, are a "special case" and need to be handled with caution.

Answer 2

It's true that most limits are computed without direct recourse to the $\varepsilon - \delta $ definition. But, of course, you must rely on some such direct computations, since otherwise you'd be making an omelette without breaking any eggs.

For instance, computing a limit of the form $\lim_ {x\to a}P(x)$, where $P(x)$, is any polynomial is very easy, you just compute $P(a)$ and you are done (since any polynomial is continuous). But how do you prove that a polynomial is continuous? Well, of course this is an easy consequence from the fact that continuous functions are closed under addition and multiplication. But what are we relying on then? Well, the fact that the constant functions are continuous and that the identity function is continuous. Easy facts, yes, but how do we prove that $f(x)=c$ is a continuous function? Well, we compute the limit at any point $a$ and show that that limit is equal to $f(a)$ (cause this is the definition of continuity). And how shall we compute the limit $\lim_{x\to a}c$? Yep, the good old $\varepsilon-\delta $ definition (how else???). Similarly, for the identity function $x\mapsto x$.

Answer 3

In the example cited in the question, $\lim\limits_{x \to 1} \ (x+4),$ it is likely to be intuitively "obvious" to a beginning student that the function $f(x) = x+4$ is continuous, which by definition implies that $\lim\limits_{x \to 1} f(x) = f(1) = 5.$ In that sense you do not "need" the $\epsilon$-$\delta$ definition of the limit, but of course there is no rigor in this approach; you intuitively assumed facts that rigor would require be proved.

Aside from rigor in cases like this, however, the $\epsilon$-$\delta$ definition of the limit is the means to prove several elementary theorems about limits, including facts about the limits of sums, products, and ratios of functions. Having proved those facts rigorously, we know they apply to all cases in which the conditions of each theorem are met, and we can happily apply them to all kinds of problems to find limits without really thinking much about the $\epsilon$-$\delta$ definition. Some of these facts, such as L'Hópital's Rule, are not nearly as intuitively "obvious" as the continuity of $f(x) = x+4$, and it is very useful to have these facts at our disposal without having to rederive them for each new function we encounter.

So that, in my view, is the practical usefulness of the $\epsilon$-$\delta$ definition to those who are not actively engaged in foundational work in mathematics. It lets the rest of us blithely continue to use those tools for finding limits that those foundational mathematicians gave us, in full confidence that they will always work as advertised.

Answer 4

No. The definition isn't really useful at all - it defines what we mean by a limit in a rigourous way, and at best can be used to prove a guessed limit really is the limit.

Many definitions are similar, in that they're not directly useful, but they are characterised nicely in other ways. Limits in particular lead to the sledgehammer that is l'Hopitals.

Answer 5

The $\epsilon$-$\delta$ definition is not at all useful for actually calculating limits. It is generally only useful in justifying limits. This is not too dissimilar from how the definition of the Riemann integral is often a terrible way to actually compute integrals.

With $\epsilon$-$\delta$ limit proofs, it is important to note that one typically knows what the limit is (or at least what you suspect it is) before writing the proof. There is no guidance in what limit to choose other than the fact that only one limit will work (if there is a limit at all).

However, sometimes intuition breaks and one does go to the definition. For example, a classic "hard problem" is to try to integrate Thomae's function (sometimes called one of Dirichlet's functions) on the interval $[0,1]$. This is the function which is $0$ at irrationals and $1/q$ at rational points $p/q$ written in least terms. This function is Riemann integrable and justification usually needs $\epsilon$-$\delta$ style formulation.

Further, Thomae's function is continuous at every irrational and discontinuous at every rational. This is not obvious, but very easy to prove.

The real value in this definition is the ability to prove abstract properties of abstract functions. One needs something precise enough to show that continuous functions are integrable at all, for instance. Similarly, to prove the fundamental theorems of calculus for continuously differentiable functions, one needs something to lean on. [You can find such a proof in some calculus texts, or any first real analysis text].

Answer 6

The $\delta-\epsilon$ definition of limits is a theorem generator for evaluating limits. Using it you can prove things like

If $\displaystyle \lim_{x\to x_0}f(x) = L$ and $\displaystyle \lim_{x\to x_0}g(x) = M$, then

$\displaystyle \quad \lim_{x\to x_0} f(x) + g(x) = L+M$

$\displaystyle \quad \lim_{x\to x_0} f(x) g(x) = LM$

$\displaystyle \quad \lim_{x\to x_0} \frac{f(x)}{g(x)} = \frac LM$ (Provided $M\ne 0$.)

You can use it to prove that all polynomials in $\mathbb R[x]$ are continuous. That composition of continuous functions is continuous (with the appropriate warnings about division by $0$)

and many more.

Answer 7

In a typical class on single variable calculus, I would imagine that you could get away without using the $\varepsilon-\delta$ definition directly and just using things like L'Hopital's rule. Certainly, when calculus was first formulated, it was done so without that rigor, and yet people were still able to get correct (and even astonishing) results. However, this is more a consequence of most familiar examples falling under a few rules than a case where a few rules cover everything imaginable.

That said, when we try to avoid the definition, we end up needing a lot of rules. For instance, if we want to know that $$\sum_{i=1}^{\infty}\frac{(-1)^i}{i}$$ converges, we want the alternating series test, but if we want to know that $$\sum_{i=1}^{\infty}\frac{1}{i^2}$$ converges, we want the integral test. Of course we can prove all these things and never worry about $\varepsilon-\delta$ again, but we end up assuming a lot of machinery which we had to construct.

In particular, when presented with a new kind of problem, we might need to go all the way back to $\varepsilon-\delta$ in order to build more machinery. For instance, consider the function: $$f(x)=\sum_{i=1}^{\infty}\frac{\sin(2^nx)}{2^n}$$ which can reasonably be seen to converge. We might wonder if $f'(x)$ exists anywhere. Termwise differentiation gives $$f'(x)=\sum_{i=1}^{\infty}\cos(2^nx)$$ which doesn't converge anywhere (since the limit doesn't go to $0$). Problem is that the termwise derivative not converging doesn't necessarily tell us anything about the function itself - for instance, we easily have that $$\sin(x)=\sum_{i=1}^{\infty}\frac{\sin(3^ix)}{2^i}-\frac{\sin(3^{i+1}x)}{2^{i+1}}$$ because the sum telescopes, but differentiating both sides (assuming that termwise differentiation works) gives: $$\cos(x)=\sum_{i=1}^{\infty}\frac{3^i}{2^i}\cos(3^ix)-\frac{3^{i+1}}{2^{i+1}}\cos(3^{i+1}x)$$ where the terms on the left go to $\infty$, so the sum doesn't converge. So, thus, we're left without tools other than $\varepsilon-\delta$ to approach such a problem (and it turns out not to be trivial).

My point here is that, yes, we usually abstract away from the $\varepsilon-\delta$ definition. We try to prove lemmas which capture our intuition and which allows us to work intuitively, yet support our work rigorously (without any mathematical contortion beyond the initial proof of whatever lemma we wanted). And, indeed, if we have an intuitive answer, we probably have some "higher level" theorem we can apply to get that result to work well with our intuition. However, given that $\varepsilon-\delta$ defines the central idea of study, sometimes we need to go back to the definition either to prove more lemmas, or to investigate some object which defies our usual approach.