Coming up with math explanations for students that don’t always “get it” can be tough. You have to be thinking on your toes, and you have to make your explanations understandable. It’s not always easy to succeed.
Two Recent Examples:
How would you explain to a student in a non-accelerated Algebra II (or even non-AP Calculus class) why it is that when you solve 
How would you explain to a student — without using the formal definition of a limit — what a limit is? So that they understand it intuitively. Now ask yourself: would your explanation work for the constant equation 
And for those of you who want to prove your mathematical mettle, here’s a question that recently circulated through our math department. What’s the answer, and how would you explain it to a stuck student?:
Suppose that a function 
Continuous Everywhere but Differentiable Nowhere 
Maybe there is a typo in the problem? It seems to me that the limit would not exist (positive infinity as h->0+ and negative infinity as x->0-).
I’m guessing that should be f(1)+h, not 1+h?
Or even f(1+h) :).
Oh Jon! You’re totally right! I have no excuse. I always get confused when I start using the /frac command in latex. Thanks!!!
Fixed above.
For the first limit question ,I’d ask the students to put it into their graphing calculators and look at the graph (changing the window to smaller and smaller parts of the domain). Then we’d do the same for the table (again, adjusting the delta x as needed).
For the log equation, again we’d go to graphing. Graph the original function and see that -3 is not in the domain.
For some reason, most of my students are able to better make the connections from the graph initially.
I’ve got nothing to help with the last one. Sorry!
For the limit problem, multiply both sides by another limit known to exist, namely lim_{h->0} h. This gives lim_{h->0}f(1+h)/h * lim_{h->0} h = 5 * lim_{h->0} h = 5*0 = 0. Since both limits on the left exist, we may combine them to get one limit and since when we take the limit h is never 0, we get:
lim_{h->0} f(1+h) = 0. Now, as f is differentiable at x = 1, it is continuous at x = 1 and hence 0 = lim_{h->0} f(1+h) = f(1). Now, we know that the derivative of f at x = 1 equals lim_{h->0} (f(1+h) – f(1))/h = lim_{h->0} f(1+h)/h = 5.
So f(1) = 0 and f'(1) = 5.
I heard this over the summer (at an AP workshop) and really liked it. We always learn about the “Learning Styles of Students” and they range from visual to naturalist and lots inbetween. But, I was often at a loss as to how it applied in my math classes. Then I was told about TWANG (I’m in Nashville, see).
T = Technological (have them graph limits or use their calc in some way to figure it out)
W = Wordy (not so applicable here, but some students actually like the words from ‘word problems’)
A = Analytical (this is the traditional rigorous way to solve problems)
N = Numerical (plug in some numbers and see if that helps give you an idea of what’s going on)
G = Graphical (look at the graph to get an idea of what to do)
So, with things like teaching logs or lims, you can figure out which style the student uses and help them along that way. If the epsilon-delta (analytic) way is too formal and confusing, then maybe go with a graphical (as Jackie says) or numerical (“plug in 0.001 and -0.001) way to help them try to understand.
@Jackie: Yup, graphing is definitely worth doing for limits. We did a ton of graphing and numerical work; we haven’t even started the analytic work. But, for the log question, for example, they’ll see graphically that there’s only one solution. But I want to answer their question: where does the second solution come in algebraically? I know the answer (domain of individual logs function is not the domain of the logs once you combine them.) But that explanation is tough — blah.
@Adam: solved it a slightly different way, but your way is much more elegant. I showed your way to the other teacher, and she thanked me, so I’m passing that along.
@David: that TWANG acronym is new to me, and I like it.
I asked my class this limit question today and one of the students came up with a rather resourceful answer. Almost immediately, she said, “if f(1) = 0, then f'(1) = 5.” I said, “but how do you know that f(1) = 0?” She said, “If your problem has a solution, then there must be only one solution. f(1)=0 doesn’t contradict anything you said, so it must be a solution and hence the only solution.”
I did point out that it is not always a good idea to trust the person who sets the problem, but I had to give her marks for cleverness. Interestingly, despite my best efforts, I couldn’t convince anyone else in the class that she was right.
When I thought it through, my way was similar to Adam’s student’s. If lim(x->0) of f(1+h) was anything other than zero then f(1+h)/h could not approach 5. It would be +/-infinity or non-existent. Since f is differentiable (and thus continuous), f(1) = 0.
An alternate method of finding f'(1) might be L’Hopital’s rule, but my guess is that it may be too early in the year for that.