Choosing $\delta$ for a given $\varepsilon$ when computing $\lim_{x\to 2}x^2=4$

Question

How does the author come to the conclusion that $\delta$ has to be the $ \min\left\{1,\frac{\epsilon}5\right\}$. Why that $1$? Can it be any number and we get accordingly the value of $\epsilon$? In general, finding $\delta$ for any $\epsilon$ is purely by some guesswork? I can understand why these values work but the ‘why’ question is troubling me. I’m not very comfortable and unable to convince myself perfectly.

Answer 1

This is one of the most annoying aspects of teaching a first course in analysis. The author of such a proof has always worked backwards, and has then rewritten the proof (erasing the entire process of coming up with it) so that it looks completely magic.

This kind of proof is generated as follows.

1. Intuit that the limit is $4$.
2. Imagining that the limit is indeed $4$, the proof will look like:

   Let $\epsilon > 0$, and choose $\delta = \mathrm{XXX}$. Pick $x$ to be within $\delta$ of $2$. Then $|x^2 - 4| < \epsilon$, which completes the proof.

3. Manipulate the final line until we can get something out of it.

   Let $\epsilon > 0$, and choose $\delta = \mathrm{XXX}$. Pick $x$ to be within $\delta$ of $2$. Then $|x^2 - 4| = |x+2| |x-2| < \epsilon$, which completes the proof.

4. Identify any terms which we can bound uniformly by picking $\delta$. This is where the $1$ comes from, and it only mattered that it was a positive constant.

   Let $\epsilon > 0$, and choose $\delta = \min(1, \mathrm{XXX})$. Pick $x$ to be within $\delta$ of $2$. Then $|x^2-4| = |x+2| |x-2| < 5|x-2| < \epsilon$, which completes the proof.

5. Repeat by successively refining $\delta$.

   Let $\epsilon > 0$, and choose $\delta = \min(1, \frac{\epsilon}{5})$. Pick $x$ to be within $\delta$ of $2$. Then $|x^2-4| = |x+2| |x-2| < 5 \cdot \frac{\epsilon}{5} = \epsilon$, which completes the proof.

6. Pretend that none of the above happened, and just write down the proof in its final form.

(Thanks to Dr Paul Russell for giving almost exactly this answer in his course Analysis II.)

Answer 2

I can relate to your difficulties. Before I answer your question, I would like to emphasize that in $\varepsilon-\delta$ proof, you need to do the "calculation" in advance.

Now, here is the thing... The concept of limit is probably best explained as moving closer and closer to a certain value. In your problem, $x$ is moving closer and closer to $2$. How close? That depends on the choice of your $\varepsilon>0$ in the first place. Now, understand that $x$ moves closer to $2$ means you can "bound" $x$. In this case, I can bound $x$ with any number (usually $1$ for simplicity and easy calculation). If you are in doubt, you can try to use another number such as $\sqrt{2}$ but maybe the calculation will be not as good sometimes.

Also, "guess work" as you mention is not entirely wrong. However, you need to be "smart" to guess the correct answer. In this case, do "rough" calculation in advance.

So, let's think in different perspective. This is our goal : $$|x^{2} - 4| < \varepsilon$$ and this is our starting point : $$ |x- 2| < \delta$$ Now, consider our goal as starting point : \begin{align*} &|x^2 - 4| = |x-2|\,|x+2| < \varepsilon \\ &|x-2| < \frac{\varepsilon}{|x+2|} \end{align*}

Now, we want to bound $x$. So, let's use different number, for example $\frac{1}{2}$. Assume $|x-2|<\frac{1}{2}$ then we have $\frac{5}{2}< x+ 2 <\frac{7}{2}$ which means $\frac{5}{2}< |x+2| < \frac{7}{2}$ and now we know that $\frac{2}{7} < \frac{1}{|x+4|} < \frac{2}{5}$

Going back to the previous inequality, we obtain $$|x-2| < \frac{\varepsilon}{|x+2|}<\frac{2}{5}\varepsilon$$

Now, we can find $\delta = \min \{\frac{1}{2}, \frac{2}{5}\varepsilon\}$ as our bound and start the proof. What I want to demonstrate here is for you to realize that "guess work" needs to be smartly done and there are a lot of answers to prove.

Answer 3

I thought I would add some reflections on what exactly is going on with $\min(1,\cdots)$. Most textbooks (and introductory classes) gloss over this entirely, and pretend it is obvious (and it is obvious for the author and anyone else who, like me, have been doing $\epsilon$-$\delta$ arguments for years). This glossing-over is a bad thing, as $\min(1,\cdots)$ is something which stumps most students when they first encounter it. They should be given a thorough reason for why we do it. Here I give something which in my opinion is a bare minimum, but about ten times more than is offered in any textbook or class I have seen (not that I have seen very many).

Note that this applies equally well to the same thing in proofs of continuity.

Limits only care about what happens very close to the point in question. In this case the point $x=2$. It doesn't care what happens at $x=1000$, for instance. Thus we care about small values of $\epsilon$ and $\delta$. So as long as $\epsilon$ is small, choosing $\delta$ to be $\frac{\epsilon}{5}$ works fine, and we're done.

However, while limits only really care about small $\epsilon$, the definition says you should be able to pick a $\delta$ for any epsilon. This is partly in order to be brief, and partly in order to be impartial as to what constitutes "small". The definition may as well have said "For any $\epsilon$ with $0<\epsilon<1$, there is a $\delta>0$ such that ..." and that would've been equivalent.

However, that's a bit longer to write, and it makes $1$ look like some kind of special number in this context, which it is not. Perhaps more importantly, some times you need to specifically use something like $\epsilon=2$ in a proof, and in that case, it's easier if the definition didn't have a limitation on $\epsilon$.

So, with that said, when we pick our $\delta$, we mostly care about small $\epsilon$, but we have to take into account that someone, somewhere, may one day pick $\epsilon=1000$. And in that case $\delta=200$ is way too big.

This can be solved in two ways. One way is to find an expression for $\delta$ which always works. That's complicated and will involve square roots, at the very least. The standard way, which is used here, is to declare that if $\epsilon$ is larger than some bound (in this case $5$), we will use $\delta=1$. If $\epsilon$ is smaller, we will use $\frac\epsilon5$.

There is nothing special about $1$ in this context. It's just that some number must be picked, and $1$ is a nice number most of the time.

Answer 4

What @arthur points out is spot on I would just like to add a couple things to it.

The reason we often choose some "maximal" delta to begin with is to make it easier to work with the piece of the function that's left. For $x^2$ at $2$ this can be nicely seen graphically.

Notice that the in the region bounded by the blue and red lines (that's between $2-1=1$ and $2+1=3$ the function $x^2$ is particularly well behaved. It is strictly increasing it is defined everywhere and so will be easy to work with. That is why we want to make sure no matter how big an $\epsilon$ someone gives us we always work in this nice region.

Doing this lets us assume nice things in our proof later on. For example $|x+2|\leq 5$ and also $x>0$. When we later try to bound $|x^2-4|$ this information can (and does) come in handy.

This is even more important when our functions are less well behaved than a quadratic. You might have functions that are not defined everywhere or behave very badly at some points. Compare to a function like $\frac{sin(1/x)}{100}$

This function is continuous everywhere except at $0$ and consequently has a limit everywhere. But if you want to prove that it has a limit somewhere other than $0$ you really want to bound yourself away from that nasty oscillation so you can work with a function that behaves nicely at the place you're taking your limit.