Math 161 Lab: The Definition of Limits  



DO NOT start Mathematica.

If you already have, please exit Mathematica before continuing.

1  The definition

Recall the definition of limit: limx® a f(x) = L means

for any e > 0,             no matter what positive number Alice picks (e)
one can find a number d > 0,              Bob can find a positive number d
such that              such that
if 0 < |x-a| < d,              if x is within d of a (but x ¹ a)
then |f(x) - L| < e.              then f(x) is as close to L as Alice specified.

Note: The Greek letters used above are called epsilon (e) and delta (d).


One can think of the definition of limits in terms of a game between two players, Alice and Bob. The game works as follows: Bob proposes a value of the limit L, Alice then challenges Bob to find an interval near a such that f(x) is within some small distance (e) from the limit L, that is Alice chooses e and challenges Bob to make sure that f(x) should be in the interval (L-e, L+e). Bob must then specify the interval of x values by providing the number d, i.e., the interval (x-d, x+d). Bob wins if every x in the interval (x-d, x+d) (except possibly x = a) satisfies f(x) Î (L-e, L + e). Alice wins if there is some x in the interval (x-d, x+d) (but not x = a) such that |f(x) - L| > e.

If the limit is indeed L, Bob will be able to find a d for any such e chosen by Alice. That is, Bob can always win. If there is some e that Alice could pick for which Bob has no winning d, then Alice can win and the limit is not L.

We can picture these intervals on a graph like the one below:

2  Applying the Definition

The Epsilon-Delta Applet provides an interactive version of this picture that we will use in this lab to explore the definition of limit a bit further. Load the applet using Internet Explorer (not Netscape). Select ``Epsilon-Delta Applet" from the list at


http://www.calvin.edu/~rpruim/courses/m161/F01/java/


Each function below has already been entered in the examples menu of the applet. Remember: Mathematica should NOT be running.

Let f(x) = 6x - x2. Consider limx ® 2 f(x).

W
hat is limx ® 2 f(x)? How does this show up on the graph?
I
f Alice picks e = 0.3 and Bob picks d = 0.1, who wins?
I
f Alice picks e = 0.03 and Bob picks d = 0.01, who wins?
I
f Alice picks e = 0.003 and Bob picks d = 0.001, who wins?
B
ased on the results above and the graphs you have looked at, if Alice picks some e > 0, what do you think Bob should pick for d? (You don't have to prove that this works, but it does as long as e is small.)
Let f(x) = x2 - 2x -1. Consider limx® 3 f(x).
S
ince f is a , L = limx ® 1 f(x) is easy to compute, namely L = . Enter this value on the second variable input line where it says ``test limit L = ". (The value of a = 3 has already been set correctly.)
I
f Alice picks e = [1/2] and Bob picks d = [1/4], who wins? Explain how you know in terms of the graph given by the applet. (You may need to zoom in.)
I
f Alice picks e = 1 and Bob picks d = [1/5], who wins? Explain how you know in terms of the graph given by the applet. (You may need to zoom in.)
I
f Alice picks e = 1, what is the largest value Bob can pick for d and still win? (Approximate this as well as you can using the graph.)
Let f(x) = x2+1. Consider limx® 0 f(x).
W
hat is limx® 0 f(x)?
I
f Alice picks e = 0.25, what should Bob choose for d?
I
f Alice picks e = 0.09, what should Bob choose for d?
S
how that if Alice picks e and Bob picks d = Ö{e}, then Bob wins. (This proves that the limit is what Bob claims it is.)
Let f(x) be the function in Example 4 of the applet. limx ® 2 f(x) does not exist. This means Alice should always be able to win the game. How should Alice pick e in order to win?

Let f(x) be the function in Example 5 of the applet. Does limx ® 1 f(x) exist? Explain.

Let f(x) = 2x sin(1/x).

W
hat is limx ® 0 f(x)?
I
f Alice picks e = 0.5, what should Bob choose for d?
I
f Alice picks e = 0.1, what should Bob choose for d?
I
f Alice picks e = 0.01, what should Bob choose for d?
I
f Alice picks some e > 0, what should Bob choose for d? Prove that your choice works.


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On 20 Sep 2001, 11:08.