# Math Help - Epsilon Delta Proof

1. ## Epsilon Delta Proof

Hi everyone.

Prove:
(x-->1)lim x^3 = 1

I have already factor x^3 into abs(x-1)*abs(x^2+x+1), but I don't know how to continue.

Thanks everyone.

2. ## Re: Epsilon Delta Proof

Originally Posted by pohkid
Hi everyone.

Prove:
(x-->1)lim x^3 = 1

I have already factor x^3 into abs(x-1)*abs(x^2+x+1), but I don't know how to continue.

Thanks everyone.
If $\displaystyle 0 < |x - c| < \delta \implies |f(x) - L| < \epsilon$ then $\displaystyle \lim_{x \to c}f(x) = L$.

You want to show that $\displaystyle \lim_{x \to 1}x^3 = 1$, so you need to show that $\displaystyle 0 < |x - 1| < \delta \implies |x^3 - 1| < \epsilon$.

Scratch work:

\displaystyle \begin{align*} |x^3 - 1| &< \epsilon \\ |(x - 1)(x^2 + x + 1)| &< \epsilon \\ |x - 1||x^2 + x + 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{|x^2 + x + 1|} \\ |x - 1| &< \frac{\epsilon}{\left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right|} \end{align*}

Now since we are making $\displaystyle |x - 1|$ arbitrarily close to $\displaystyle 0$, we can restrict it to be no more than a certain distance from $\displaystyle 0$, say $\displaystyle |x - 1| \leq 1$.

\displaystyle \begin{align*}|x - 1| &\leq 1 \\ -1 \leq x - 1 &\leq 1 \\ 0 \leq x &\leq 2 \\ \frac{1}{2} \leq x + \frac{1}{2} &\leq \frac{5}{2} \\ \frac{1}{4} \leq \left(x + \frac{1}{2}\right)^2 &\leq \frac{25}{4} \\ 1 \leq \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} &\leq 7 \\ \left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right| &\leq 7 \end{align*}

So $\displaystyle \frac{\epsilon}{\left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right|}$ will be at its minimum when $\displaystyle \left|\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}\right| = 7$.

So we can let $\displaystyle \delta = \min\left\{1, \frac{\epsilon}{7}\right\}$ and reverse the process, and you will have your proof.