Limits

By Mark Roulo

Last Updated: 9-Feb-2020

Introduction

This is not an introduction to Calculus, nor a tutorial.

This is, instead, a collection of mental errors that I think students often make and an attempt to correct them. This is a work in progress, but as of now, two big points:
  1. Notation is Misleading
  2. Don't Think in Terms of Infinitesimals

Notation is Misleading!

Something that must be clear up front is that the notation used when performing Calculus encourages incorrect mental models. The mathematical community does this to save time when writing out the math. The cost is that vast numbers of students think they know what is happening, but are wrong about that.

So ...

A) ∞ is Not a Number!

The following two expressions both use limits and look identical, but they are not:
lim x → 10

and:
lim x → ∞

The first thing to notice is that 10 is a number and ∞ is not.

The "problem", such as it is, is that one can get 'closer' to 10, but one cannot get closer to infinity. So the first expression means:

What is the limit as we consider x values closer and closer to 10?

while the second expression means

What is the limit as we consider x values that get bigger and bigger with no limit ... ?

You need to think about these (at least a little) differently, and it is probably better to start thinking about limits that "go to" infinity before thinking about limits that "go to" a given constant (interestingly, G. H. Hardy's "A Course in Pure Mathematics" does exactly this). Beginning with limits going to constants such at zero or ten encourages incorrect mental models about infinitesimals that will break down later.

...

Again, two different things: The first makes sense (but is wrong).
The second (fairly obviously) doesn't make sense ...

B) δ is Not a Number, Either!

C) dy dx Looks like division, but it isn't!

And, in fact, typical Calculus books do mention that this is neither division nor a ratio.

The problem is that the students have spent about 10 years treating this notation as division, so mentioning ONCE that THIS use of the notation is different doesn't make much of an impression. Possibly worse is that treating this like division sorta works, so the student is told (once) to think differently about something that looks familiar, but then things sorta work when the student doesn't. This is a recipe for misunderstanding.

And this become EXTRA important when the chain rule is introduced and used:

dy dx = dy du ⋅ du dx

Because the students want to think that this is happening:

dy dx = dy du ⋅ du dx

This is NOT what is happening! d/dx is an operator (like "+" and "√") and there is no "cancellation". More accurate notation is:

dy dx = d du y ⋅ d dx y

Or, even more explicitly:

dy dx = ( d du y ) ⋅ ( d dx y )
OPERATIONAL RECOMMENDATION: The students should write their solutions using the notation that makes the operator-ness explicit.

Don't Think in Terms of Infinitesimals!

Consider the (first) fundamental theorem of Calculus

d dx f(x) = lim h → 0 f(x+h) - f(x) (x + h) - (x)

And lets consider a popular f(x) used to illustrate:

f(x) = x 2

So, what is the derivative with respect to x of f(x)=x2?

A solution that gets the right answer — "2x" — but with a mistaken view of what is happening is something like this:

StepMath Student Thinking
1) d dx f(x) = lim h → 0 f(x+h) - f(x) (x + h) - (x) Original problem
2) = lim h → 0 (x + h) 2 - (x) 2 (x + h) - (x) Plug in x2 for f(x).
3) = (x 2 + 2xh + h 2 ) - (x) 2 h I remember my algebra!
4) = 2xh + h 2 h More algebra!
5) = h ⋅ (2x + h) h Let me factor a bit!
6) = h h ⋅ (2x + h) And I'll rearrange ...
7) = (2x + h) I can drop hh because h ≠ 0 so hh is 1.
8) = 2x And here I can drop the h because h = 0.
If h was ≠ 0, then 2x ≠ 2x + h and the results would be different.

The symbol manipulation in the last two steps is correct-ish, but the student thinking is very wrong.

Part of the problem is that the 'h' in the limit here looks like a real number (which is what all the algebra was about), but it isn't!

And ...

lim h → 0   h h ⋅ (2x + h) ≠ h h ⋅ (2x + h)

The "mistake" above starts in step (3) when the student dropped the lim h → 0 . Dropping the limit changes the problem! [Is this correct? The limit is what happens to the result, not to the input!]

Basic Idea

The basic idea

Continuity: No Hopping Around!

If the value is "hopping around" then there isn't a limit, even if the value is heading towards a specific value.

Removable Discontinuity

(Removing the hole ...)

Jump/Step Discontinuity

Infinite Discontinuity

Discontinuous Everywhere

Consider this function:

f(x) = { 0 , x ∈ ℚ 1 , x ∉ ℚ

The way to read this is that f(x) is 0 if x is rational (e.g. 1, 2, ⅓) and f(x) is 1 if x is irrational (e.g. π, e). A graph of the function looks something like this (with the bit at y=0 raised a tiny bit so that the x-axis line doesn't obscure it):
x y
This function has no limits anywhere.

Now, how about this function:
f(x) = { 0 , x ∈ ℚ 1/x , x ∉ ℚ

Four Basic Types of Problems

These tend to come out of the discontinuities:

Evaluating at number

There is a 'hole' in the function at number

What happens at ∞

The function heads to ∞ at number