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re:[apcalculus] 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 yvalues 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 xcoordinate 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 ycoordinate 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  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus To unsubscribe click here: http://lyris.collegeboard.com/read/my_forums/ To change your subscription address or other settings click here: http://lyris.collegeboard.com/read/my_account/edit



