Play the

Limit Game

Adam claims that f(x) approaches a number L when x approaches the number a, i.e. he claims

but Beth doubts that. How can we establish who's right? They have to play the limit game. If both play perfectly, then the winner is right. The game is played in three rounds:

Beth, the doubter has to establish what she is willing to accept to be "close to L". She tells us a small number e > 0 of error tolerance. In other words, all numbers between L-e and L+e are considered to be "close to L".
Next Adam, the claimer has to tell what he considers to be "close to a" by giving a tolerance d > 0. All numbers between a-d and a+d are now considered to be "close to a".
Finally it's Beth's move again. She has to pick a sample number x "close to a", i.e. a number between a-d and a+d, but different from a.
Both compute f(x) and check whether it is "close to L", i.e. whether it is between L-e and L+e. If it is, Adam (the claimer) wins and the above statement is true, otherwise Beth (the doubter) wins and the statement is false.

Window Version

There is also a shorter version, but involves a more sophisticated test. Whereas in the above version it suffices to compute f(x) for some number and check whether it is within two bounds, now the test consists to plot the graph and check the shape of this graph.

Beth chooses e > 0.
Adam chooses d > 0.
Both check whether the graph of f in the window defined by a-d < x < a+d and a-d < y < a+d enters this window on the left, and leaves on the right, without leaving the window at top or bottom.

Towards the e-d Definition of Limits

When we are about to claim a limit like

we have to be able to win against all possible moves of all possible Beths. What we need is a strategy that allows us to respond to every possible e of the first step. Our choice of d in step 2, the only choice we make during the game, must of course depend on this e. Obviously we counter small es with small ds. You may consider e to be the independent variable influencing d, the dependent variable. If we have found such a function d = g(e) which provides us for every positive e with a successful (in the remaining steps) positive d, then we can be sure to win the limit game no matter how Beth plays.

Example: We want to claim

What strategy would work? Try to reflect Beth's e. So choose d: = e. How close must f(x) be to 6 if x is d-close to a? We look at the deviation |f(x)-6| = |(x²-9-6x+18)/(x-3)|= |(x²-6x+9)/(x-3)| = |(x-3)²/(x-3)|. Since x must be different from 3, this is equal to |x-3|. Since x is d-close to a, this is smaller than d, i.e. smaller than e, as required.

All limit laws can be derived in this way.

Limit Laws

Assume that both

then

		sum law
		constant multiple law
		product law
	if M š 0	reciprocal law
	if M š 0	quotient law (follows from reciprocal and product laws)

Proof: We use the above e-d method. in all cases we delegate the problem of finding the appropriate d for any given e for f respectively g to two subcommittees.

Then how could these subcommittees help us to find an appropriate d for e for the sum function? Well, we receive Beth's e but communicate instead e/2 to each subcommittee. They provide us with positive numbers d₁ and d₂ such that |f(x)-L|<e/2 for every x d₁-close to a, and |g(x)-M|<e/2 for every x d₂-close to a. We choose d to be the smaller of d₁ and d₂. Then. for every x that is d-close to a, we have both |f(x)-L|<e/2 and |g(x)-M|<e/2, therefore we get |f(x)+g(x)-(L+M)|<e, as desired.

How would the subcommittees help us in the other cases? For the constant multiple law, obviously we only need the first subcommittee, and we would communicate e/2 to it, and take the d the subcommittee would provide.

In the product law case, note that |f(x)g(x)-LM| = |f(x)g(x)-f(x)M+f(x)M-LM| = |f(x)(g(x)-M)+(f(x)-L)M| Ł |f(x)|*|(g(x)-M)|+|(f(x)-L)|*|M|. Try to find out what you send to the subcommittees, and what you do with the received d₁ and d₂ by yourself.

Give also a proof for the reciprocal law.

Do we need a proof for the quotient law?

Erich Prisner, October 2003