Interesting but very easy epsilon-delta problems?

Question

I am teaching a real analysis class. Students in the class have inconsistent high school algebra skills. They now have a complete but tenuous understanding of $\varepsilon$-$\delta$ limits. I want to give them problems on which to consolidate their understanding of this definition and how to prove that a given proposed limit satisfies it. I want more examples to try out with them that are interesting but do not overly tax their algebra skills or demand too much sophisticated inequality-based reasoning, which is entirely new to them. I am looking for problems. Answers to this question could consist of problem suggestions, or pointing me to sources.

To calibrate what I'm looking for, I will go in detail through an example I tried with them last class that was too hard. (It is not very hard if you have certain analysis skills, but these are precisely the skills they don't have.)

We worked through an example like $\lim_{x\to -5}10x$; they felt they understood everything that was going on. I threw $\lim_{x\to 4} 3x^2$ at them, thinking it was just one level up, but it was way too much. They attacked it like this:

The limit is 48, by substitution. We need to find the constraint on $x$ that causes $|3x^2 - 48|$ to be $<\varepsilon$. So we need

$$-\varepsilon < 3x^2 - 48 < \varepsilon$$ i.e. $$-\varepsilon + 48 < 3x^2 < \varepsilon + 48$$ i.e. $$ -\frac{\varepsilon}{3} + 16 < x^2 < \frac{\varepsilon}{3} + 16$$ and so $$ \sqrt{-\frac{\varepsilon}{3} + 16} < x < \sqrt{\frac{\varepsilon}{3} + 16}.$$

Well and good (subject to $\varepsilon \leq$ something), but now they needed a $\delta$ such that $0<|x-4|<\delta$ will force this last inequality. Using a taylor series approximation for the square root is not something they would have ever thought of. I proposed working with the previous inequality prior to the square root. My idea was to try to get them to square $4-\delta < x < 4 + \delta$ and then choose $\delta$ to force $-\varepsilon / 3 + 16 < (4-\delta)^2 < x^2 < (4+\delta)^2 < \varepsilon / 3 + 16$. First of all, I had to hand them this move. Secondly, finding the necessary $\delta$ was still hard, once I had handed them this setup. They went:

$$-\frac{\varepsilon}{3} + 16 < (4 - \delta)^2 = \delta^2 - 8\delta + 16$$ so we need $$ -\frac{\varepsilon}{3} < \delta^2 - 8\delta$$ or $$ 8\delta - \delta^2 < \frac{\varepsilon}{3}$$

and at this point started trying to remember how to solve a quadratic equation. I stopped them and pointed out that $\delta^2$ is positive, therefore this inequality is guaranteed as soon as $8\delta < \varepsilon / 3$, and they immediately concluded that $\delta$ had to be $<\varepsilon / 24$.

Then we looked at the right inequality, $(4 + \delta)^2 < \varepsilon/3 + 16$. They got as far as

$$\delta^2 + 8\delta < \frac{\varepsilon}{3},$$

and again I had to hand them a trick: pointing out that it never hurts to take a smaller delta, we can assume $\delta \leq 1$, implying $\delta^2 \leq \delta$, so that $\delta^2 + 8\delta \leq 9\delta$. From here, they concluded we need $\delta < \varepsilon / 27$. So in the end they chose $\delta =\min(1,\delta / 28)$.

Because of all the tricks I had to hand them, and handling the sides separately, it was hard for them to see the big picture at once, so in the end, I don't feel the problem consolidated their understanding of the definition.

What I want is several problems that are not all the same (i.e. not just a bunch of $\lim_{x\to c} f(x)$ for $f(x)$ linear) but that do not put up the types of algebraic/analytic roadblocks I described above: places where you either have to (a) do some slightly heavier HS algebra, like solve a quadratic and then reason clearly about how its solution plays into these inequalities, or (b) use the analyst's trick of exploiting the slack afforded by the inequalities, as above when I pointed out to them we could take $\delta \leq 1$ in order to have $\delta^2 \leq \delta$, which is its own separate skill I would prefer to work on with them once we already have the essential goals of the $\varepsilon$-$\delta$ proof mastered.

Again, I am asking either for your own suggested problems, or suggestions about good sources for such problems. Thanks in advance.

ADDENDUM: In addition to Brendan's nice suggestion of degree 1 rational functions, I found some other nice ideas in Exploratory Examples for Real Analysis, by Snow and Weller:

• Piecewise-linear functions, where you're taking the limit at a joint.

• Things like $(x-5)\sin x$ as $x\to 5$, where the key is just to notice that $|\sin x|\leq 1$.

I haven't tried them with my students yet to see how they will work, but these are just the types of ideas I was looking for.

Answer 1

I suggest using rational functions. Students are used to evaluating limits of rational functions because such examples are prevalent in most calculus courses. Moreover, I think the work required to prove such a limit is a little more challenging than that for a linear function but mostly in terms of the algebra and a little bit of logic. But, by following the thought process I outline below, I think students can learn to see the choice of $\delta=\min\{blah,yadda\}$ as a natural, intuitive choice based on their understanding of limits, and not some mystically dark mathemagic.

Claim: $\lim_{x\to 0}\frac{2}{x+1}=2$

Scratch work for proof: Let $\varepsilon>0$ be arbitrary and fixed. The goal is $2-\varepsilon < \frac{2}{x+1} < 2+\varepsilon$ so we'll work backwards from there. We want to eventually see something of the form $|x|<\delta$ since $x\to 0$ is the limit.

Multiply through by $x+1$ everywhere: $2x+2-x\varepsilon-\varepsilon < 2 < 2x+2+x\varepsilon+\varepsilon$

Did we just multiply through by a negative and need to switch the inequality signs? Well, no, as long as $x+1>0$. In other words, as long as $x>-1$, we're fine. Since $x\to 0$, the intuition is that $x$ is eventually close to 0, and so certainly greater than $-1$. So, let's leave this aside for now and worry about it later if we need to.

Subtract 2 from everywhere: $2x-x\varepsilon-\varepsilon<0<2x+x\varepsilon+\varepsilon$

Factor out the $x$ where we can, since we're hoping to see $|x|<\delta$ sometime soon: $x(2+\varepsilon)-\varepsilon < 0 < x(2+\varepsilon)+\varepsilon$

We also want to see $x$ in the middle, so let's subtract that factored $x$ expression everywhere: $-\varepsilon < -x(2+\varepsilon) < \varepsilon$

Divide by $(2+\varepsilon)$ everywhere: $-\frac{\varepsilon}{2+\varepsilon} < -x < \frac{\varepsilon}{2+\varepsilon}$ (Did we just divide by a negative? Surely nope, since $\varepsilon>0$ so $2+\varepsilon>2$.)

Now we have something like $|x|<\delta$. Regardless of $x$'s sign, as long as its magnitude is less than $\frac{\varepsilon}{2+\varepsilon}$, then we have what we want.

So, we would like to set $\delta=\frac{\varepsilon}{2+\varepsilon}$. Then, we could start with $|x|<\delta$ and work all of these steps backwards to get to $\left|\frac{2}{x+1}-2\right|<\varepsilon$

However, there was that one iffy step when we needed $x>-1$. When writing our proof, working that particular step backwards, we would be dividing through by $x+1$ everywhere. At that point, all we would know about $x$ is that it's between $-\delta$ and $+\delta$. What if $\delta$ were too big? What if $\delta \geq 1$? Then it's possible for $x\leq -1$. Oops! How do we avoid this?

The only way that would happen is if the $\varepsilon$ we were given made it so that $\frac{\varepsilon}{2+\varepsilon}\geq 1$. However, algebraically solving this yields $0\geq 2$, so that is impossible. Indeed, $2+\varepsilon>\varepsilon$, so the ratio must be smaller than 1. That's it! The $\delta$ we choose will work for any $\varepsilon$.

You could even point out here that, if we hadn't noticed whether or not $\frac{\varepsilon}{2+\varepsilon}\geq 1$ were possible, we could have just effectively said, "Make sure $\delta\geq 1$ is never possible. If our formula, in terms of $\varepsilon$, would make that happen, instead just make $\delta$ something smaller, like $\frac12$." This motivates a definition like: $\delta=\min\{\frac12,\frac{\varepsilon}{2+\varepsilon}\}$. In this example, you don't even need to do that, but you can use it as an occasion to point out why you might in other situations.

(Final note: I initially misread your post as requesting examples of limits with sequences, taking an $\varepsilon$ and choosing a sufficiently large $N$. I based the above example on a sequence example I used in my Real Analysis course just last week: $\lim_{n\to\infty}\frac{2n}{n+1}=2$. I converted that example into the one above by setting $x=\frac{1}{n}$. You can do this to create your own similar examples.)

Addendum: I do like user Santiago Canez's comment about bounding $|x|$ versus bounding $|x+1|$. I suggest presenting that after this method and having a discussion with the students comparing the two. Ultimately, the same proof is written, but it will be instructive for them to compare the different modes of thinking to create that proof.

Answer 2

What I would in this situation as instructor is to rewrite $3x^2$ for them as follows:

$3x^2=48+24(x-4)+3(x-4)^2$

Although this will seem like pulling a rabbit out of hat to students, verification is simple algebra.

From here, students should be able to focus on the analysis involved in estimating $|3x^2-48|$ in terms of $|x-4|$.

Answer 3

You are making this WAAAAAY too complicated. One simplification is you should be centering around the value you're taking the limit at. So in this case, if you define d = x-4, then x = 4+d and 3x² = 3(16+8d+d²). Since we want an upper limit, for small d we can replace d² with 8d, so we have 48+48d. So if we want that within $\epsilon$ of 48, we just have to have d<$\epsilon$/48. Simple.

This is generally how to do these proofs:

1. Define d = x-x₀

2. Find D(d) = f(x₀+d)-f(x₀)

3. Find an upper bound B(d) on D(d) that's easily invertible.

4. Set $\delta$ = B^-1($\epsilon$)

If the function has a nice derivative, the process is even simpler. The derivative of 3x² is 6x. Take a neighborhood around 4, say 0 to 10, and the derivative is at most 60. So this gives $\delta = \epsilon/60$.

So while this isn't an answer to your question "How do I get simpler problems?", I think this is the wrong question to be asking, because the problem isn't that your problems are to complicated, it's that your techniques are too complicated. I don't see much point in asking "What are some problems that can still be solved even if you're going about them completely the wrong way?"