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I'm revisiting the materials I've put together for students taking a non-proof-based intro to calculus, and my goal is for them to have a clear but rough sense of a limit as a bound (basically enough of a genuine understanding for the limit definition of the derivative to have actual meaning to them; but the focus of the course will very much be using calculus as a tool).

I've started off by introducing limits via tables of values (using simple functions, to establish the idea of "approaching-but-not-equal") and would like to next move to direct substitution (and then gradually illustrate the limitations of direct substitution); however, I'm a bit at a loss as to how to introduce direct substitution.

The texts I've consulted all introduce direct substitution by invoking continuity in one way or another (i.e. directly or indirectly), but this is a bit too ouroboros for my liking (insofar as it just muddies the water to invoke continuity to justify limits, when the only definition of continuity they'll encounter uses a limit -- and only later on, at that).

At this point, I'm leaning towards something as vague as "for many functions we can ... introduce substitution", -- any suggestions for a cleaner introduction / justification? or is this just not possible in the scope of an intro course for "applied calculus"?

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    $\begingroup$ Can you say why you want to introduce computing limits by direct substitution at such a stage, and why you want to display its limitations only gradually? If I were a student, I think I would really start to wonder why we bother with limits at all if all I were seeing were examples that could be done by direct substitution. Better, in my opinion, to start with examples where direct substitution doesn't work, so that students have a clearer motivation for studying limits. $\endgroup$ Dec 29, 2021 at 20:28
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    $\begingroup$ At any rate, I think there's no way to avoid mentioning continuity in talking about direct substitution. If you really don't want to postpone discussing direct substitution until after continuity, just state that you are using the intuitive notion of continuity, which will be more rigorously defined later using limits and which will then give an after-the-fact justification for direct substitution. $\endgroup$ Dec 29, 2021 at 20:33
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    $\begingroup$ Keisler's calculus book was recently mentioned. (I've not seen it!!) Further research by Kathleen Sullivan $\endgroup$
    – Rusi
    Dec 30, 2021 at 3:39
  • $\begingroup$ Found the other question -- contains link to keisler online. And other links matheducators.stackexchange.com/questions/5989/… (Yeah not exactly recent...) $\endgroup$
    – Rusi
    Dec 30, 2021 at 4:33
  • $\begingroup$ @Raciquel, do you mean "what does $f$ approach as $x$ approaches $c$?" $\endgroup$ Dec 30, 2021 at 7:11

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I think that it makes sense to introduce continuity in the same lesson that you introduce limits.

Here is a sketch of a lesson plan:

Give them this link

https://www.desmos.com/calculator/rlu2zgcjyf

Group work:

  1. Is $f(2)$ defined or undefined?
  2. As $t$ approaches $2$, what are the values of $f(t)$ approaching?
  3. Make a table of values for $t = 1.9, 1.99, 1.999$ and $t=2.1, 2.01, 2.001$.

Summarize and introduce the words and notation for limits. Stress the connection between the numerical and graphical approaches.

Then show them this related graph:

https://www.desmos.com/calculator/zxivpgtrfl

Point out that, in this graph $g(2)$ exists and is equal to the limit which had been calculated before. Define continuity at a point.

Now reveal the formulas:

The formula for the first function is $f(x) = \frac{x-2}{x^2 - 3x + 2}$. Plug in $f(1.99)$ as a sanity check (and to reinforce the numerical/symbolic connection).

Point out that attempting to plug in $x= 2$ results in $\frac{0}{0}$, which is undefined. If we make $x$ close to $2$, but not equal to $2$, then we will get a number close to $0$ divided by a number close to $0$. Spend some time here. If you have clickers you could ask some multiple choice questions like "$\frac{0.0000001}{7}$ is (a) extremely small, (b) extremely large, (c) neither". Then repeat for $\frac{7}{0.0000001}$, $\frac{0.0001}{0.0000001}$, $\frac{0.0000001}{0.0001}$, and $\frac{0.0002}{0.0001}$.

Summarize the findings: when investigating comparing fractions making the numerator smaller makes the fraction smaller, but making the denominator smaller makes the fraction larger. In the case we are looking at there is a "tug of war": both are approaching zero. Which of these drives will win? As the last three numerical examples show in the last paragraph, the answer depends on the relatives sizes of the two quantities, which is hard to predict by just looking at the formula.

Question: can anyone simplify the formula $\frac{x-2}{x^2-3x+2}$?

Answer: Someone gets $g(x) = \frac{1}{x-1}$. Point out that this function is identical to the one we started with, expect it is defined at $x=2$. Plug in $g(1.99)$ as a sanity check. In other words, this is the formula for the second function we graphed.

Can we tell the limit of this function as $x \to 2$ easily? Yes! If $x$ is getting close to $2$, it makes sense that $\frac{1}{x-1}$ is getting close to $\frac{1}{2-1} = 1$.

In other words, we have found $\displaystyle \lim_{x \to 2} f(x)$ by using algebra to find another function $g$ so that $f(x) = g(x)$ when $x \neq 2$, but $g$ is continuous at $x=2$. So finding $\displaystyle \lim_{x \to 2} g(x)$ is easy: just plug it in (direct substitution).


The work I wish I could require for these kinds of problems reflects this way of thinking.

In my ideal world, here is what a student response to the following limit problem would look like:

Find $\displaystyle \lim_{x \to 4} \frac{x-4}{\sqrt{x} - 2}$.

The function $f(x) = \frac{x-4}{\sqrt{x} - 2}$ has a domain of $x \in [0,4) \cup (4, \infty)$. Since $f$ is not defined at $x=4$, it is not continuous at $x=4$.

When $x$ is close to $4$, the formula for $f$ gives me something of the form $\frac{\textrm{close to 0}}{\textrm{close to 0}}$, which is indeterminate. So I will need to do more work to determine the limit.

However,

$$ \begin{align*} f(x) &= \frac{x-4}{\sqrt{x} - 2}\\ &= \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x} - 2}\\ &= \frac{1}{\sqrt{x}+2} \end{align*} $$

Thus $g(x) = \frac{1}{\sqrt{x}+2}$ is a new function which is continuous on $[0, \infty)$ and agrees with $f$ everywhere except for $x=4$.

So

$$ \begin{align*} \lim_{x \to 4} f(x) &= \lim_{x \to 4} g(x)\\ &= g(4)\\ &= \frac{1}{\sqrt{4}+2}\\ & = \frac{1}{2+2}\\ &= \frac{1}{4} \end{align*} $$

So $\displaystyle \lim_{x \to 4} \frac{x-4}{\sqrt{x}-2} = \frac{1}{4}$.


I will also note that this kind to analysis pinpoints why so many instructors would be uncomfortable with the work:

$$\displaystyle \lim_{x \to 4} (x+3) = \displaystyle \lim_{x \to 4} 7 = 7$$

even though all of the equalities are correct.

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  • $\begingroup$ On the off-chance someone casually reading doesn't see your point at the end regarding the equality $\displaystyle \lim_{x \to 4} (x+3) = \displaystyle \lim_{x \to 4} 7,$ I can't think of a reasonable justification for the equality that doesn't essentially rely on knowing $\displaystyle \lim_{x \to 4} (x+3) = 7,$ and so the equality doesn't really qualify as an explanatory step towards justifying $\displaystyle \lim_{x \to 4} (x+3) = 7.$ Thus, while the equalities are true, I bet the writer could not formulate them in this order as a "two-column proof". $\endgroup$ Dec 30, 2021 at 16:47
  • $\begingroup$ @DaveLRenfro I agree. The principle we want to use to justify all of these limit equalities is really "two functions which are equal on a punctured neighborhood of the limit point have the same limit at that limit point". The algebraic manipulations which are allowed are those which obey this. The whole processes is to manipulate the expression until we find one which defines a continuous function on an interval containing the limit point. $\endgroup$ Dec 30, 2021 at 17:16
  • $\begingroup$ Thanks for the thoughtful & detailed reply. Is it fair to summarize your approach as "illustrating direct substitution and equivalence of limits using a graphical example and then relying on students' intuition to extend it to other functions"? Or do you take additional steps to justify the use of direct substitution on other functions? Also, in your answer you say "... $g$ is continuous ... " but it's not clear to me when / how you are explaining that $g$ is continuous (from the above outline, I'd be concerned students would conclude it means "not having holes in the domain"). $\endgroup$
    – Rax Adaam
    Dec 30, 2021 at 19:53
  • $\begingroup$ @RaxAdaam Sure, it is hard to summarize one (or multiple!) classes in one post here. I would introduce the concept of a limit, shortly after introduce the definition of continuity, and I would say that every function which is built out of compositions of addition, subtraction, multiplication, division, trig function, logs and exponentials are all continuous on their domains. The only way to write "formulas" which are discontinuous involve piece-wise defined functions or functions whose domains have holes. $\endgroup$ Dec 30, 2021 at 21:29
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    $\begingroup$ @RaxAdaam I introduce limits first, then continuity. Limits are not defined, but motivated (I don't want to do $\epsilon-\delta$). Continuity is then defined in terms of limits. Since there is no definition of a limit, the "theorems" about continuity cannot be proven, and must be taken on faith. No one questions that, for example, the sum of two continuous functions is continuous though. $\endgroup$ Dec 31, 2021 at 0:02

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