Expert: Ahmed Salami - 9/21/2010

Question
1. The problem statement, all variables and given/known data
prove the statement using epsilon delta definition of the limit

lim x->2 (x^3)=8
2. Relevant equations
lim x->a f(x)=L is true when
|f(x)-L|<e
whenever
0< |x-a|<d


3. The attempt at a solution

|x^3 - 8| < e
|(x-2)| |(x^2+2x+4)|<e
x^2+2x+4 has no real roots there for only has one sign f(0)=4 so it is always positive
so |x^2+2x+4|=x^2+2x+4
|(x-2)||(x^2+2x+4)| <C(x-2)<e
where
|(x^2+2x+4)|<C
(x-2) < (e/C) =d

I usually solve most problems like this but I can't find the C
I know that we usually say that d is a small distance so |x-a| <1
|x-2|<1
1<x<3
but then what exactly?  

Answer
Hi Hamad,
From 1 < x < 3,
2 < 2x < 6 and 1 < x² < 9
Therefore,
1+2+4 < x² + 2x + 4 < 9+6+4
7 < x² + 2x + 4 < 19
or simply |x² + 2x + 4| < 19
Hence
|x³ - 8| < 19|x - 2|   if |x - 2| < d ≤ 1
But
19|x - 2| < e   if |x - 2| < e/19
So, if we take d as the smaller of 1 and e/19, then
|x³ - 8| < e   if |x - 2| < d
which then proves the limit.

Regards