Preliminary Analysis
We want to make abs((x^2-3x)-(-2)) < epsilon∣∣(x2−3x)−(−2)∣∣<ε
By choosing deltaδ, we control the maximum size of abs(x-2)|x−2|
We now look at abs((x^2-3x)-(-2))∣∣(x2−3x)−(−2)∣∣ keeping in mind what we control.
abs((x^2-3x)-(-2)) = abs(x^2-3x+2)∣∣(x2−3x)−(−2)∣∣=∣∣x2−3x+2∣∣
= abs((x-1)(x-2))=|(x−1)(x−2)|
= abs(x-1)abs(x-2)=|x−1||x−2|
If we knew the size of abs(x-1)|x−1|, we could choose abs(x-2)|x−2| to make sure that the product is < epsilon<ε
Let's start by making sure that xx is a little close to 22, say abs(x-2) < 1|x−2|<1 (deltaδ will need to be at most 11.)
If abs(x-2) < 1|x−2|<1, then -1 < x-2 < 1−1<x−2<1, that is: 1 < x < 31<x<3.
If follows that 0 < x-1 < 20<x−1<2, so we have abs(x-1) < 2|x−1|<2
Now if we ALSO make sure that abs(x-2) < epsilon/2|x−2|<ε2,
then we will have abs(x-1)abs(x-2) < 2abs(x-2) < 2 (epsilon/2) = epsilon|x−1||x−2|<2|x−2|<2(ε2)=ε
Now we are ready to write the proof:
Proof
Given epsilon > 0ε>0, let delta = min{1,epsilon/2}.
(Note that #delta is positive.)
For every x that satisfies 0 < abs(x-2) < delta
we have
(abs(x-1) < 2 and also)
abs((x^2-3x)-(-2)) = abs(x^2-3x+2)
= abs((x-1)(x-2))
= abs(x-1)abs(x-2)
< (2)(epsilon/2) = epsilon
We have shown that for any epsilon > 0 there is a delta > 0 such that
if 0 < abs(x-2) < delta, then abs((x^2-3x)-(-2)) < epsilon.
By the definition of limit, lim_(xrarr2)(x^2-3x) = -2.
