For proving that

Delta epsilon proof: limit x->2 (x^3 - 3) = 5

The following proof is given:

Given e > 0 we need to find d > 0 such that

|x - 2| < d ==> |(x^3 - 3) - 5| = |x^3 - 8| < e.

One way to do this is to expand x^3 - 8 into powers of (x - 2).

x^3 - 8 = [(x - 2) + 2]^3 - 8 = (x - 2)^3 + 6(x - 2)^2 + 12(x - 2).

Then, by the triangle inequality,

|x^3 - 8| = |(x - 2)^3 + 6(x - 2)^2 + 12(x - 2)|

<=|x - 2|^3 + 6|x - 2|^2 + 12|x - 2|.

Now, we insist on d <= 1. In this manner,

|x^3 - 8| <=|x - 2| + 6|x - 2| + 12|x - 2|.

= 19 |x - 2|.

Finally we can fully specify d. Given e > 0, let d = min{1, e/19}.

Then, |x - 2| < d ==>

|x^3 - 8| <= 19 |x - 2|

< 19 (e/19)

= e, as required.

I'm lost as to how the exponents can be dropped as seen in the bolded text. Anyone have an idea?