1. Feb 3, 2006

### 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

2. Feb 4, 2006

### 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:
$$x ^ 3 - (-8) < \epsilon$$, 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?
$$|x ^ 3 + 8| < \epsilon$$, you need |x + 2|, so let's factor it.
$$\Leftrightarrow |x ^ 3 + 2 ^ 3| < \epsilon$$
$$\Leftrightarrow |(x + 2) (x ^ 2 - 2x + 4)| < \epsilon$$
$$\Leftrightarrow |x + 2| \times |x ^ 2 - 2x + 4| < \epsilon$$, everything is positive, divide both sides by |x2 - 2x + 4|
$$\Leftrightarrow |x + 2| < \frac{\epsilon}{|x ^ 2 - 2x + 4|} = \frac{\epsilon}{|(x - 1) ^ 2 + 3|} < \mbox{what ?}$$.
Can you go from here?