When we're defining a limit, we use (epsilon) and (delta) as distances along x- and y-axis. Epsilon and delta are infinitesimal distances.

Soon or later in the course, you will be given some statements and you need to prove that lim _{x® a} f(x) is equal to the number L by using epsilon-delta definition. I am sure that your instructor will prove some statements. But you need to prove them by yourself to fully understand and know how to prove them. Depending on what the function f you're dealing with, you will encounter some difficulties. But most functions would be nice polynomials, so don't you worry too much. I am 90% sure that this epsilon-delta proof will be on your first exam, so be sure to know how to prove this by heart.

The following example will demonstrate Definition 1.1.1

Example 1.1.1:

Prove lim _{x® 0} x = 0.

Solution:

You can right away tell that this statement is true since lim x® 0 x is equal to 0. But in mathematics, even though it looks simple and obvious, you need to prove it.

There are actually 2 parts in proving epsilon-delta definition:

1. To guess a value for epsilon
2. To Show that delta works

Part 1: You need to let epsilon be a given positive number.

In here a=0 and L=0, so you need to find delta such that

lx-0l< eps whenever 0< x < del.

So x < eps whenever 0 < x < del.

Now, pick del=eps.

Part 2: Since eps > 0, you can let del=eps.

If 0 < x < del, then x < del = eps.

So lxl < eps implies lx-0l < eps.

Therefore, limit is equal to 2.

The preceding example is one of the simplest example that I can come up with.

So the proof is very short. In an exam, you are lucky to have this kind of proof. From my experience, most instructors would not give a proof this easy on their exams. If your instructor were to prove this, he/she may have a different proof from mine. Your instructor or other instructors will have a slightly different style of proving this. It is just a style that each person prefer.

Problem 1.1.1:

Prove by using eps-del definition of limit.

Lim _{x® 2} (x² + 2x + 1) = 9

Problem 1.1.2:

Prove by using eps-del definition of limit.

Lim _{x® 0} lxl = 0

*epsilon = eps and delta = del.