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Topic: [ap-calculus] Limits
Replies: 2   Last Post: Aug 7, 2011 11:58 AM

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 Dave L. Renfro Posts: 2,165 Registered: 11/18/05
re:[ap-calculus] Limits
Posted: Aug 7, 2011 10:44 AM

Teresa Jackson wrote (in part):

http://mathforum.org/kb/message.jspa?messageID=7516569

> I want to know what is the best way to explain the following
> to one of my students. This is my first year teaching AP
> Calculus AB.
>
> I gave the following two problems to my students for homework:
>
> 1. Find the limit of f(x) = sin (pi * x) as x approaches 1.
> Some answered zero and the others answered DNE (does not exist)
>
> 2. Find the limit of f(x) = cos (1/x) as x approaches 0.
> All answered DNE

censor foil

> As a class discussion, we said that problem #2 doesn't have
> a limit because the function is oscillating infinitely and
> it never approaches a single height. One student said that
> the y-values are bouncing between negative and positive values
> as you approach zero from the left and right and the function
> never settles on any one value.
>
> For problem #1, a student asked why does the limit exist for
> this function because the sine graph is an oscillating function.

One possible method for #1 is substitution. Letting u = pi*x,
we're looking at the limit as x approaches 1 of sin(u), which
is equivalent to the limit as u approaches pi of sin(u). In the
last equivalence, I'm making use of the fact that x --> 1 implies
u --> pi, AND u --> pi implies x --> 1 (i.e. the operation "x --> 1"
is equivalent to the operation "u --> pi"). Note that both
implications are needed, which you can see by considering an
example such as exp(-1/x^2) as x --> 0 replaced with, via u = x^2,
exp(-1/u) as u --> 0. Of course, now students have to know that
the limit of sin(u) as u approaches pi is equal to sin(pi) = 0,
and I don't know how they're supposed to REALLY know this without
essentially knowing that sin(x) is continuous at x = pi.

However, assuming this is mostly an informal approach to limits,
here's one way to handle it. From a graphical standpoint, look
at the behavior of the graph as you mentally trace along the
graph towards the point with x-coordinate 1. Does the graph
"come together" at a point as you trace towards the point
from both sides? (Of course, it is irrelevant whether the
"point of togetherness" is a point on the graph.) If YES,
then the limit exists and the limit is equal to the y-coordinate
of the "point of togetherness". If NO, then the limit doesn't
exist. In this particular example, it might also be helpful
to review the fact that the graph of y = sin(pi*x) is simply
a horizontal stretch of the graph of y = sin(x). Yes, I know
this won't be a review for some students and others will have
totally forgotten it, so it provides an opportunity to squeeze
in a few minutes of precalculus review in a setting relevant
to them, so it will more likely stick with them.

In the case of #2, all the wiggles are squished into the
vicinity of x = 0, so the graph doesn't "come together"
there. Of course, you'll also want to consider the example
y = x*sin(1/x) as x approaches 0, which also has infinitely
many wiggles squished into the vicinity of x = 0, leading
you to refine the explanation of y = sin(1/x) by saying that
the wiggles don't damp out as you approach x = 0. Then there
are the examples y = f(x) and y = x*f(x), where f is the
function that equals 1 when x is rational and 0 when x is
irrational, which don't really have wiggles but they seem
similar to sin(1/x) and x*sin(1/x), leading to a more precise
(but also more abstract) idea of "function fluxuations"
damping out or not damping out as you trace towards a point.
(And then maybe introduce the squeeze theorem for limits
if this digression goes on long enough ...)

Dave L. Renfro
====
Course related websites:
http://apcentral.collegeboard.com/calculusab
http://apcentral.collegeboard.com/calculusbc
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