Deltas and Epsilons

The Formal Definition of Limit

Teachers are often concerned that the Advanced Placement Calculus tests do not test the formal definition of limit and the delta-epsilon idea. I think there are two reasons for this.

  1. For linear functions the relationship is always \delta =\frac{\varepsilon }{\left| m \right|} where m is the slope of the line. Anyone can memorize that, so testing it does not provide any evidence that the student knows anything. For any other function (polynomials of higher degree, trigonometric functions, etc.) some special one-time “trick” is required, often different for each type of function. Thus, the question becomes unreasonably difficult even for one who understands the concept.
  2. As a multiple choice question, since “none of these” is never a choice on AP tests, the smallest answer must be correct. If \delta =\tfrac{1}{2}\varepsilon  is correct, then so is \delta =\tfrac{1}{3}\varepsilon ,\ \delta =\tfrac{1}{10}\varepsilon  etc. Again no real knowledge of the definition is required – just choose the smallest delta.

Of course, the definition of limit is important. You may teach it in your class. It just will not appear on the AP Calculus exams.

Here are some notes The Definition of Limit contributed by Paul A . Foerster of San Antonio, Texas, that will help your students understand the concept.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s