"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.

BACKGROUND/OVERVIEW
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!

DEVELOPING THE PROBLEM (SCAFFOLDING)
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=1
n=3,m=1
n=3,m=2
n=1,m=2
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.

FOLLOW-UP: THE GENERAL CASE
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!

0.99999.. equals 1: Oh no, not another 'Proof!'

For the remainder of this post, the statement 0.99999... = 1 will be denoted by S.

Over the course of my math education and my professional teaching career, S has occupied considerable time and provoked much thought on my part and reflection among my students, countless mathematicians and, now, the math blogosphere (see Polymathematic's famous series of posts!). Sane individuals (aka, non-mathematicians) remain skeptical about S, unwilling or unable to grasp the equality in the statement.
They argue: "0.99999... gets closer and closer to 1 but how can you say it EQUALS 1. There's always a gap!" Ah, the mystery of limits!

For many years now, I have been posting my 'proof' of S on various listservs, discussion groups (including MathShare, the one I moderate) and blogs. Here's the reaction I 've generally received: ____________________
That's right - silence. Because I like to put a positive spin on things, I take that to mean no has found a way to refute it! I've even occasionally heard a student say that this convinced her/him.

I don't want to bore the veterans out there who've heard and read all of the well-known arguments, most of which have 'holes' in them (or should I say, discontinuities!). Even using the basic formula for the sum of an infinite geometric series doesn't necessarily satisfy the Odd Thomases (sorry, I'm a Dean Koontz addict) who will continue to question the validity of the statement.

Any attempt to justify S necessarily requires (to paraphrase Liping Ma) a profound understanding of fundamental principles regarding the real number system and my argument is no different.

Enough already -- Here it is:

Non-Rigorous Explanation: If 0.9999... is less than one, then there must be a decimal between it and 1. But this is impossible!

Rigorous Explanation:

Step 1: Consider the sequence: 0.9, 0.99, 0.999,...
Since this is an increasing sequence of real numbers bounded above by 1, this sequence has a limit, L, namely its least upper bound. As many of you know, I am using the Completeness Axiom for the Reals (known by other names). An excellent reference for the axiomatic structure of the real number system can be found here.

This demonstrates that 0.99999... does exist (i.e., it is a real number). Thus,
0.99999... is the limit L of the above sequence. Verification of the existence of 0.99999... is what is often lacking in other demonstrations of S.

Step 2: L is either greater than 1, equal to 1 or less than 1. We need only consider the last 2 cases.

Step 3: Reasoning indirectly, assume that L<1. By the density property of the real numbers, there must exist at least one real, x, between L and 1. Since L is different from x, it must differ from it in some decimal place. The tenths place? No! Since x is less than 1 and greater than L, it must have 9 in the tenths place. The hundredths place? No, again for the same reason. Need I continue or do you see we've reached a contradiction? Therefore, our assumption that L is less than 1 is false. Thus, L = 1 or, equivalently, 0.99999... = 1. QED!

Ok, your turn! Feel free to critique the proof or present your own favorite argument for or against S. Also, would you consider using this type of argument when teaching this topic?

Totally Clueless Challenge #2 - By All Means!

It's been awhile but something this good is always worth waiting for!
TC has sent me some fascinating challenge problems for our readers. If you are now sick of watching amateur videos on the Arithmetic and Geometric Mean Inequality, it's time to raise the bar. The following involves a well-known generalization of these means but the results are worth your efforts, particularly parts (c) and (d) below.

If a and b are positive, we can define their generalized mean to be:

GNM = ((a^k + b^k)/2)^(1/k)

This would look far prettier in LaTeX but I'm hoping it's readable. In words, we're looking at:
The kth root of the arithmetic mean of the kth powers of a and b.

(a) What is another name for the result when k = 1? (we're starting off easy here!)
(b) What is another name for the result when k = -1? (slightly harder algebraically)
(c) Ok, now for the real challenge for you Calculus lovers:
What is the limit of GNM as k-->0? The result is totally cool!
(d) TC's Super Bonus: Show that the limit of GNM as k-->∞ is the maximum of a and b.

Note: These have been slightly edited from tc's original problems, but they are essentially the same. Solutions may be posted in a couple of days although the notations will be hard to render. I might just have to do another video or wait for that special technology I mentioned earlier! We're hoping some of you will tackle the harder ones and comment!

Back to Your Roots: Investigating Nested Radicals