Calculus I—Day 4

Thanks to a Thursday holiday, we finished our first week of calculus after four lectures. I’m not sure that I would want it any other way—we had time to ease into the material, work out technical kinks with the software, and work on some homework assignments without too much pressure.

What I Taught

Since there had been some confusion, I spent the first 15 minutes of class trying to give students another way of thinking about limits. I showed them, using Maple, how one can think of a function approaching a limit near a point \(a\) if, when the \(y\)-range is fixed on a graph, the \(x\)-range can be “zoomed” to a small enough range so that the entire graph of the function near \(a\) can be made to fit on the screen, without going off the top or bottom.

From here, I picked up right where the previous lecture left off, and introduced another tool for computing limits using continuity. Specifically, I noted that limits can be moved into compositions of functions (given some mild hypotheses), and we used this to prove that \(f\circ g\) is continuous at \(a\) if \(f\) is continuous at \(g(a)\) and \(g\) is continuous at \(a\). I finished the section by stating the Intermediate Value Theorem.

The rest of the lecture was devoted to examining limits at infinity. The definition was stated in a manner parallel to “ordinary” limits, and I noted that the limit laws apply to limit at infinity without much modification (for instance, the limit of a sum is the sum of the limits, and so on). We had some time for examples at the end of class, so, in a refreshing change of pace from the generally frantic nature of the class, we did some actual computations for 20 minutes before taking a quiz.

What Worked

I was really rather proud of the demonstration of limits using Maple. There were definitely some students for whom the idea of a limit really clicked when they saw the examples. I am filing this under “What Worked” because I think that it did work for almost everyone, but it is a qualified success, as there were still one or two students who seemed really confused (maybe even more confused than before). On the other hand, limits are a difficult concept, and it takes time to get used to them.

I am also happy that we finally had the time to really work through some examples. In the fall and spring, this class is taught with three hours of large-group lecture every week, plus two hours of small-group recitation. The smaller recitation sections are used to do computations, review for exams, and to give quizzes. They allow for more one-on-one interaction, and are an invaluable practice.

Unfortunately, the compressed summer class keeps all of the lecture and eliminates the recitation sections. On the plus side, the lecture sections are kept smallish, which makes some small-group interaction possible from time to time. On the minus side, there is little opportunity to work on examples, which means that I largely have to rely on the Math Center and homework assignments to give students an opportunity to work through some computations. This is a mixed bag, at best, as many students don’t really take the homework seriously, or don’t make time to seek outside help.

Hence working examples in class counts as a major win. Woohoo!

What Didn’t Work

I always try to have five or six examples ready to go before every lecture, so that if there is time, we can work through a few. In general, I write the examples to slowly ramp up the amount of theory that is required, and introduce new ideas one at a time. Sometimes I skip around a bit to highlight important ideas when there is not enough time. When I started working examples, I managed to get through the first two, noticed that time was getting short, and skipped to the last one. Turns out that this was a bad idea.

The last example involved several algebra “tricks” and technicalities that the first two examples didn’t require. To make matters worse, I had not put a step-by-step solution into my notes, so I ended up forgetting a crucial detail in the computation which really muddied the waters as we worked through the problem.

The problem was as follows: compute
\lim_{x\to-\infty} \sqrt{4x^2+2}+2x.
The first thing to try is a simple application of the limit laws, which renders \(\infty-\infty\), which I know to be nonsense. However, the naive answer is that anything minus itself is zero, hence the limit should be zero. This logic is incorrect, but the students had not seen a counter example, and this problem wasn’t meant to provide one (even assuming that a later mistake hadn’t made things really confusing).

That said, I managed to convince them that \(\infty-\infty\) didn’t make sense, and we tried something else. In this case, there is a somewhat magical rationalization of the “numerator” that makes things work out. That is,
\lim_{x\to-\infty} \sqrt{4x^2+2}+2x
= \lim_{x\to-\infty} \left[\sqrt{4x^2+2}+2x\frac{\sqrt{4x^2+2}-2x}{\sqrt{4x^2+2}-2x}\right]
= \lim_{x\to-\infty} \frac{2}{\sqrt{4x^2+2}-2x}.
This step comes out of nowhere for most of the students, and was the primary new idea that was meant to be introduced by this example. However, because of the confusion about \(\infty-\infty\), I don’t think that this got the emphasis that it needed and deserved.

Finally, there is another subtlety that I simply lost track of. The next step is to move the \(x\) terms into the denominators of fractions, as \(\lim_{x\to\pm\infty} 1/x = 0\), which allows for easy simplification. This is generally done as follows:
\lim_{x\to-\infty} \frac{2}{\sqrt{4x^2+2}-2x}
= \lim_{x\to-\infty} \left[\frac{2}{\sqrt{4x^2+2}-2x}\frac{1/x}{1/x}\right]
= \lim_{x\to-\infty} \frac{2/x}{\sqrt{4+2/x^2} – 2}.
Or, at least, this is generally how it is done if you are not paying attention. Can we spot the error? It is subtle.

In general, something like \(x=\sqrt{x^2}\) holds true. However, there are issues of sign. Note that \(\sqrt{x^2} = |x|\), which is always a positive number, hence if \(x\) itself is negative, there is a sign error. Since the limit above is at negative infinity, \(x\) is negative for all interesting values of \(x\), which means that \(x=-\sqrt{x^2}\). So the correct simplification is
\lim_{x\to-\infty} \frac{2}{\sqrt{4x^2+2}-2x}
= \lim_{x\to-\infty} \frac{2/x}{-\sqrt{4+2/x^2} – 2}.

The first computation leads to \(0/0\), which is a problem. The second computation correctly leads to the solution
\lim_{x\to-\infty} \sqrt{4x^2+2}+2x
= \frac{0}{-4}
= 0,
which is the correct result. However, by the time I got to the incorrect answer of \(0/0\), I had less than 10 minutes to administer a quiz, so we finished on a rather confusing note. I’ll have to revisit that problem again during the next lecture and try to fill in the gaps. I’ll make sure to put the full computation in next time.

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