Can someone show me how to solve this?
Prove limit[x^3, x, 2]=8
First of all, I have no idea what CAS speak is. Second, all the information I have is what I wrote. I am taking this as a correspondence course so I have no professor and write things the way I see them. I'm sorry if I'm doing something wrong. There are no calculus classes anywhere near me that I can take and this is my only way. This is a required class or I would not even be taking it in the first place.
You want to show $\displaystyle \lim_{x\to 2} x^3 = 8$.
Let $\displaystyle |x-2| < \delta$ then it means $\displaystyle ||x|-|2||\leq |x-2| < \delta \implies 2 - \delta< |x| < 2 + \delta$.
In addition let us require that $\displaystyle 0 < \delta \leq 1$ thus $\displaystyle |x-2|<\delta \implies 1< |x| < 3$.
Now $\displaystyle |x^3 - 8| = |(x-2)(x^2+x+4)| = |x-2||x^2+x+4| \leq |x-2|(|x|^2+|x|+4|) < |x-2|(9+3+4)$
Thus, we get that $\displaystyle |x^3-8| < 16|x-2| < 16\delta$.
With this fact we can complete the proof. Let $\displaystyle \epsilon > 0$. And choose $\displaystyle \delta = \min (1, \tfrac{\epsilon}{16} )$.
Then by the above reasoning if $\displaystyle 0 < |x-2| < \delta = \tfrac{\epsilon}{16}$ then $\displaystyle |x^3-8| < 16\delta \leq 16\cdot \tfrac{\epsilon}{16}= \epsilon$.