# Understanding Limits Using Epsilon-Delta Proofs

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Sometimes we have to evaluate the limit of a function for a value that is undefined for the function. In these cases, we can explore the limit by using epsilon-delta proofs.

## Epsilon-Delta Proofs

Your friend gives you two locks, which are linked together. Having a key will let you unlink the locks. Links between function limits and x values work the same way.

In a limit statement, the value we are approaching is linked to how we evaluated the function at this value. Sometimes, the value we are approaching is undefined by the function. In these cases, we can unlock the concept of the limit by using epsilon-delta proofs.

For example, let's look at a particular function defined for all values of x except at x = 5. At x = 5, this f(x) is undefined. For f(x) = 4x, this is a straight line with a hole in it.

We would like to know f(x) as x gets close to 5. In other words, we are looking for the value of 4x in the limit as x goes to 5. We will call this value 'L'. There is a concise way to say all of this. We want to find the limit L so that:

Remember, x = 5 is not an allowed value for this function. Otherwise, we would simply evaluate f(x) at x = 5.

## The Epsilon-Delta Proof

To 'unlock' this limit concept, we use the epsilon-delta proof.

We explore values for x between 5-δ and 5+δ. The δ (read this Greek letter as 'delta') is some positive number. We could write this with inequalities as:

5-δ < x < 5+δ

We could make this even more compact by subtracting 5 from each term:

• 5-δ - 5 < x - 5 < 5+δ - 5
• -δ < x - 5 < δ
• Using absolute value signs: | x - 5 | < δ

## Absolute Values

Is | x - 5 | < δ really the same as -δ < x - 5 < δ?

That's the beauty of the absolute value sign. For example, if we had | x | < 2, then we could write: -2 < x < 2.

Check out some positive and negative values for x:

• If x = 1, then -2 < 1 < 2 and | 1 | < 2.
• If x = -1, then, -2 < -1 < 2 and | -1 | is +1, which is less than 2.

To recap, | x - 5 | < δ is the same as -δ < x - 5 < δ, which means 5 - δ < x < 5 + δ.

## A Range of Values for f(x)

We start with an ϵ (read this Greek letter as 'epsilon'), which is a number greater than zero and a value of δ. This defines values for x, which give a f(x) within an interval from L - ϵ and L + ϵ. Using inequality signs we can write:

L - ϵ < f(x) < L + ϵ.

Saying L - ϵ < f(x) < L + ϵ is the same as saying -ϵ < f(x) - L < ϵ, which is the same as | f(x) - L | < ϵ.

To recap, saying

means choosing how close we want to be to L by specifying an ϵ with | f(x) - L | < ϵ. We then choose our x value so that | x - 5 | < δ.

The idea behind the epsilon-delta proof is to relate the δ with the ϵ. Let's do this for our function f(x) = 4x.

If L were the value found by choosing x = 5, then f(x) would equal 4(5) = 20. We know f(x) itself does not exist at x = 5 but the limit may exist.

• Start with | f(x) - L | < ϵ to get: | f(x) - 20 | < ϵ.
• Substituting for f(x), we get: | 4x - 20 | < ϵ.
• This gives us: -ϵ < 4x - 20 < ϵ.
• Divide through by 4: -ϵ/4 < x - 5 < ϵ/4.
• Compare -ϵ/4 < x - 5 < ϵ/4 to -δ < x - 5 < δ. See the x - 5 in both expressions?
• We conclude: δ = ϵ/4.

If we choose an ϵ to get close to L, we can calculate ϵ/4 and pick an x between 5 - ϵ/4 and 5 + ϵ/4.

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