# Thread: Lost on Cubic Epsilon Delta proof

1. ## Lost on Cubic Epsilon Delta proof

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?

2. It's because $\displaystyle |x - 2| \leq \delta$ and you are making $\displaystyle \delta \leq 1$.

So $\displaystyle |x - 2| \leq 1$.

Therefore $\displaystyle |x-2|^3 \leq 1^3 = 1$ and $\displaystyle |x-2|^2 \leq 1^2 = 1$.

Also note that $\displaystyle |x - 2|^3 \leq |x - 2|$ for $\displaystyle |x - 2| \leq 1$ and $\displaystyle |x - 2|^2 \leq |x - 2|$ for $\displaystyle |x - 2| \leq 1$, since exponentiating a fraction between $\displaystyle (0, 1)$ makes the fraction smaller.

So $\displaystyle |x - 2|^3 + 6|x - 2|^2 + 12|x - 2| \leq |x - 2| + 6|x - 2| + 12|x - 2|$.

3. Thanks. I get the whole inequality thing now. All I'm lost on now is the theoretical aspect of it. What allows us to take the 19 and divide E by it such that d = E/19.

I understand the logic behind doing so for 2nd degree polynomials. For example, when proving that the limit of X^2 +1 = 5 as x approaches 2, I understand why we take X - 2 < d, set d at 1, add to all sides to get 3 < X +2 < 5, and then replace E/|x+2| with E/5.

I'm not sure what we are doing here. Factoring it all out, I can now get the answer right for any cubic equation because you just take the coefficient. But I'm not sure why exactly we are allowed to divide E by the coefficient like we divide E by 5 above.

4. You have gotten to the point that

$\displaystyle |x^3 - 8| \leq 19|x - 2|$ as long as $\displaystyle \delta \leq 1$.

Also recall that $\displaystyle |x - 2| \leq \delta$.

So surely $\displaystyle |x^3 - 8| \leq 19\delta$.

Remember that you were trying to find an $\displaystyle \epsilon$ such that $\displaystyle |x^3 - 8| < \epsilon$. Surely here you can let $\displaystyle \epsilon = 19\delta \implies \delta = \frac{\epsilon}{19}$, as long as $\displaystyle \delta < 1$.

So we say that $\displaystyle \delta := \min \left\{1, \frac{\epsilon}{19} \right\}$.