Friday, September 19, 2008

"Fun" With Limits Early in Calculus

Update: The limit below can easily be derived using L'Hopital's theorem. The purpose of this article is to provide practice in algebra and limit manipulations, limit properties and the definition of the derivative prior to using this theorem.

Important Notes:

(1) The condition n ≠ 0 can be relaxed. At some point, students should be asked to analyze the need for restrictions.
(2) The instructor may well want to avoid giving students the above formula, preferring to have them derive it at least in the positive integer case (see comments below in red under "Developing the Problem").
(3) Note that m and n are not restricted to be positive integers. It is recommended that the instructor begin with this restriction on m and n to allow for an algebraic derivation.
(4) This is not an introductory limit exercise! Please read comments (red, bold) below about starting with concrete numerical values before attempting this generalization.

This is the time of year when Calculus students quickly move into those wonderful limit problems. Epsilon-delta arguments may not be as popular these days but the mechanics of limits are still the challenge for students. Those who have taught this know that students generally struggle with the algebraic simplifications and procedures. Other than these manipulations students generally feel this topic is easy:

Possible Student Thinking: "You just do some algebra, eliminate the "bad" factor in the denominator and plug in. Easy stuff!"

Naturally, if the assignments contain more theoretical limit problems, they may not feel that way!

On the other hand, the algebra can be a major stumbling block for the more challenging exercises. In this post, I will uncharacteristically deemphasize the theory behind the "cancel and plug in" technique and focus on the algebra at first. Then we will move on to relating the limit to the definition of the derivative and application of some important limit properties, in other words, theory! Of course, if L'Hopital's Theorem were introduced early on (in the chapter on differentiation), that would clearly be the method of choice for students!

If m and n in the limit above are positive integers, students can attempt to factor out "x-a" from the numerator and denominator and substitute x = a into the resulting reduced expression. However, many students struggle with such general factoring formulas (or may not have seen them.) Therefore, synthetic division can be used to generate the other factor. This reviews some nice Algebra 2 but what if m and n are not positive integers? What if they are rational or even irrational? Standard factoring techniques would not apply in general so what to do?

I certainly am not suggesting that the instructor begin with the general problem. In fact, I would 'concretize' the problem using a few special cases:
n=2,m=2 (This special case is worthwhile as it reviews basic definitions and limit properties).
n=4,m=5 (requires more sophisticated factoring or synthetic)

Based on these exercises, the instructor may ask students if they can develop a general formula for any positive integer exponents. this is in lieu of giving them the formula at the beginning.

After the definition of the derivative is given, students can attempt the more general version. This is a fairly sophisticated limit manipulation but one worth assigning. I may outline the method in an addendum to this post or in the comments or wait for one of our astute readers to contribute! As a hint, the technique I used is related to the derivation of L'Hopital's Theorem!


Totally_clueless said...

In my calculus classes, one of the early limits discussed was

Limit (x->a) (x^n - a^n)/(x-a), which is n.a^(n-1)

I assume this is discussed in current calculus classes also.

Can one use this to tackle the problem, i.e., Decompose it as

(x^n-a^n)/(x-a) . (x-a)/(x^m-a^m)

and just multiply the limits.

Does the general approach apply to limit problems: can we multiply the function by another whose limit goes to 1 and attempt to simplify the problem, or is there a pitfall that I have overlooked?


Dave Marain said...

Nice derivation using the derivative. As far as my understanding of the theory goes (I always defer to the experts here), the original expression and the product of the two expressions you wrote are identical for x ≠ a. Therefore, if each separate limit exists (and they do), we can apply the Product Rule for Limits to justify this rigorously. This is quite similar IMO to the actual proof of L'Hopitals Theorem for f(x)/g(x) in which the numerator and denominator each evaluate to zero at x=a. We then rewrite the expression as ((f(x)-f(a))/(x-a))/((g(x)-g(a))/(x-a)
and apply the definition of the derivative and the quotient rule for limits as before. This proof however may require more rigor involving non-zero derivatives, etc.

One of my goals here was to initially provide additional practice for algebraic manipulation in limits which shows up early on in the course. For example, the limit as x-->a of (x^3-a^3)/(x^5-a^5). Students forget how to factor the difference of cubes and most likely have not seen the general form for x^n-a^n. Therefore, they would need synthetic division by x-a to do this or Mathematica or the TI-89 or some Symbolic Algebra Manipulator. Of course, once they recognize hte "rewrite" strategy you employed, they might go that way.

In the end, how do they react when we show them L'Hopital's Rule, typically at the end of the semester: "Why didn't you show us that earlier!" Right, like we're going to show them the short-cut first. Like any of us ever learned that way!

The other piece here is the derivation when m and n are not positive integers. This does not allow for simple algebraic manipulation...

Eric Jablow said...

The interesting limit question is not the one of taking takes the limit as x goes to a. It's the question where m/n goes to an arbitrary real number t.

Dave Marain said...

Please elaborate, Eric. Are you referring to rational approximations to an irrational or something else? I should probably think about this more after I get some rest!

Eric Jablow said...

You want to consider the case where m/n is irrational. Note that replacing m by cm, n by cm, and x and a by their c-th roots change none of the expressions here. So, for m/n rational, this can be considered a function of m/n.

Let t be an arbitrary irrational, and write it as the limit of rationals. Show that lim_{m/n->t} lim_{x->a} (x^m-a^m)/(x^n-a^n) can make sense if you swap the limiting expressions, with that rescaling as necessary. It's a somewhat vague thing to do, but it's similar to what one does in calculus with the chain rule to find dx^t/dx for any x:

Suppose t = m/n. Let y = x^t. Then y^n = x^m.

ny^{n-1}y' = mx^{m-1}. Substitute for y to get y'. Use limiting arguments on t to show that the formula holds for irrational t.

Dave Marain said...

Nice Eric! In the back of my mind, I thought you might be deriving the irrational case using a double limit argument.

Eric Jablow said...

A better way to think of it is: Let b = a^n and y=x^n. Also let r=m/n Then,

lim_(y->b} (y^r -b^r)/(y-b) = lim_{x->a} (x^m - a^m)/(x^n - a^n) = m/n a^(m-n) = r b^(r - 1).

Now, let r go to irrational t. Can swapping the limits in

lim_{r->t} lim_{y->b} (y^r - b^r)/(y - b)

be justified?