# Functions whose derivatives can be found without knowledge of limits

I'm teaching a precalculus course. One of the topics covered in the course is the difference quotient. The textbook simply presents it as an alternative expression of the average rate of change of a function and leaves it at that. That's fine, but I feel like many students (for good reason) don't see the 'point' of the difference quotient in a precalculus class.

For that reason, for this semester, I'm presenting the difference quotient as a way to calculate instantaneous rate of change. However, since this is a precalculus course, they have no knowledge of derivatives and limits.

I'm basically going to present problems in which they calculate instantaneous rate of change of $f(t)$ at $t = t_{0}$ by:

(I) Calculating the difference quotient

(II) "Simplifying fully"

(III) Letting $h=0$ and $t= t_{0}$.

My question is, which types of functions will this work for? Does it work for all polynomials? Does it only work for polynomials? Is there a name for such functions (where $h$ can be set to $0$ to yield the derivative rather than calculating the limit)?

Thank you.

• Letting $h=0$ would likely seem unjustified, without some discussion of limits. My recommendation is to first introduce limits. The informal concept of limit, and some of the simpler calculations of limits are often done in Precalculus courses. – quasi Aug 26 '18 at 14:54
• Well it works for polynomials IF the students are comfortable with the binomial theorem (which seems a bit unlikely). And with a little algebra you can handle functions like $\sqrt x$ this way. But, really, you need the notion of a limit to justify these operations anyway so... – lulu Aug 26 '18 at 14:56
• Teaching too much non-precalculus material as part of a standard Precalcus course is unfair to the students, in my opinion. Giving some introduction to limits is fine, but I would leave it at that. Let the rest of the Calculus material wait until they take Calculus. Why? Because there's so much actual precalculus material, and mastery of that is key to the student's future success in math. – quasi Aug 26 '18 at 15:10
• I agree with the others saying not to do this, but if you must, there are plenty of such computations you can find in Stack Exchange, such as Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ and Differentiation using first principles with rational powers. And look at the "Related" and "Linked" questions at each of these. – Dave L. Renfro Aug 26 '18 at 15:54
• Also, if you feel you must do this (and if you want questions they can't easily google or look up solutions), perhaps you can have them evaluate limits of symmetric difference quotients $\frac{f(x+h) - f(x-h)}{2h}$ and various "lopsided" difference quotients, such as $\frac{f(x+3h) - f(x-2h)}{5h}$ and $\frac{f(x+2h) - f(x-h)}{3h}.$ For functions that are differentiable in the usual sense, these limits will exist and be equal to the derivative. – Dave L. Renfro Aug 26 '18 at 15:58

There's an answer to your question, sort of, below. But first: Please don't do this! It's simply wrong, mathematically (unless you do it just right, as in the "answer" below - but doing that seems if anything more complicated than simply talking about limits before derivatives). We both know that for most of the students it's not going to matter much what you do - otoh for the students who are later going to learn some calculus from a mathematically correct perspepctive, you're giving them wrong notions that they're going to have to unlearn. Like "division by zero is not allowed, except when you're doing calculus".

Calculus was nonsense mathematically for a long time; coming up with a mathematically valid version was one of the great accomplishments of Cauchy, Weierstrass et al. Ignoring that and reverting to the bogus version seems very sad.

(Also, not that this is the important issue, but it amounts to feeding the trolls. There are crackpots and web sites out there that assert that calculus is simply wrong. Typically the reason given is an explanation of why setting $h=0$ is invalid. Of course that's not a refutation of actual calculus, because in actual calculus we do not set $h=0$. But...)

My point, in the unlikely event that it's not clear: Say $f(x)=x^2$. If $h\ne0$ then $$\frac{f(x+h)-f(x)}h=2x+h.\tag{i}$$But now setting $h=0$ and concluding $f'(x)=2x$ is wrong, because (i) is only true for $h\ne0$.

You wouldn't even have to mention the word "limit" to give an unobjectionable version of the argument. Say (i) holds for $h\ne0$, and deduce that if $h$ is very small then the left side of (i) is very close to $2x$.

If someone asks why you don't just say "set $h=0$" give him or her a gold star and explain why it's not quite correct to put it that way.

Now, about the question of for what functions the procedure works: Whether it works is really not a property of the function so much as a property of the formula used to represent the function. In an abstract sense it works for any differentiable function. Here's an easy lemma:

Lemma. The function $f$ is differentiable at $x$ if and only if there exists a function $g(h)$, defined near the origin and continuous at the origin, such that $(f(x+h)-f(x))/h=g(h)$ for small $h\ne0$; in this case $f'(x)=g(0)$.

So. That function $g(h)$ such that $f'(x)=g(0)$ always exists, the question is whether you can find a formula for $g(h)$, $h\ne0$, such that the formula defines a function continuous at the origin.

Not that I'm saying you should do things in class from that point of view - surely that introduces more baggage than just talking about limits. My point is just that in some sense "it always works". For example if $f(x)=\sin(x)$ and $x=0$ then $$g(h)=\begin{cases}\frac{\sin(h)}h,&(h\ne0), \\1,&(h=0).\end{cases}$$Of course that doesn't actually help in finding the derivative, because there's the question of why setting $g(0)=1$ makes $g$ continuous at the origin.

But the function $g$ such that $f'(0)=g(0)$ is out there - the reason this doesn't "really work" is we can't write down a simple formula for $g(h)$ that's valid for $h\ne0$ and also for $h=0$.

• For an example (that the OP may find of interest) of a difference quotient calculation in which plugging in $h=0$ is defined but doesn't give you the derivative, see Paramanand Singh’s answer to Why doesn't derivative difference quotient violate the epsilon-delta definition of a limit? (the lower half “Update” part). – Dave L. Renfro Aug 26 '18 at 16:15
• @DaveL.Renfro Yes, that's a good example. The last sentence, "plugging does not work in above example because the greatest integer function $\lfloor x \rfloor$ is not an elementary function" is not quite right, or is at least missing the point: The reason it doesn't work is that the floor function is not continuous. – David C. Ullrich Aug 26 '18 at 16:34
• Regarding the "elementary function" comment, I had not noticed this before, which incidentally wasn't in my original 15 March 2006 ap-calculus post (cited in a comment there). I've left a comment there for P. Singh to perhaps reword this part. – Dave L. Renfro Aug 26 '18 at 17:54