Comments on: ‘Splanations

By: David P

David P — Thu, 09 Oct 2008 01:56:16 +0000

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.

By: Adam Glesser

Adam Glesser — Thu, 09 Oct 2008 01:34:53 +0000

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.

By: samjshah

samjshah — Thu, 09 Oct 2008 00:14:37 +0000

@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.

By: David P

David P — Wed, 08 Oct 2008 21:04:05 +0000

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.

By: Adam Glesser

Adam Glesser — Wed, 08 Oct 2008 04:39:04 +0000

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.

By: Jackie

Jackie — Tue, 07 Oct 2008 11:02:02 +0000

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!

By: samjshah

samjshah — Tue, 07 Oct 2008 10:45:58 +0000

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.

By: Jon Ingram

Jon Ingram — Tue, 07 Oct 2008 08:16:39 +0000

Or even f(1+h) :).

By: Jon Ingram

Jon Ingram — Tue, 07 Oct 2008 08:15:41 +0000

I’m guessing that should be f(1)+h, not 1+h?

By: David P

David P — Tue, 07 Oct 2008 03:08:45 +0000

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-).