Mastering Limits: Proving f(x) = x^3 --> -8 as x --> -2 | Step-by-Step Guide

[elle]

Hi,

I'm currently working on a proof problem and I'm just totally clueless on what I've been asked to prove.

The question is:

Define the real valued function f by f(x) = x^3 and prove that f(x) --> -8 as x --> -2. [ use the epsilon and delta definition of a limit ]

So far this is my attempt but I'm stuck where I'm trying to find epsilon. I'm not even sure my attempt is even correct Can someone help? Thanks in advance!

http://www.tinypic.com/view/?pic=n49nqv"

 

[VietDao29]

Now you are looking for a \delta in terms of \epsilon, such that:
If 0 < |x - (-2)| = |x + 2| < \delta then |x ^ 3 - (-8)| < \epsilon, for some given \epsilon > 0, right?
Now work backward, assume that:
x ^ 3 - (-8) &lt; \epsilon, we will try to rearrange it to make it look like: |x + 2| < something, and then we can let \delta = \mbox{something}, and finish our proof. Do you get it?
|x ^ 3 + 8| &lt; \epsilon, you need |x + 2|, so let's factor it.
\Leftrightarrow |x ^ 3 + 2 ^ 3| &lt; \epsilon
\Leftrightarrow |(x + 2) (x ^ 2 - 2x + 4)| &lt; \epsilon
\Leftrightarrow |x + 2| \times |x ^ 2 - 2x + 4| &lt; \epsilon, everything is positive, divide both sides by |x² - 2x + 4|
\Leftrightarrow |x + 2| &lt; \frac{\epsilon}{|x ^ 2 - 2x + 4|} = \frac{\epsilon}{|(x - 1) ^ 2 + 3|} &lt; \mbox{what ?}.
Can you go from here?