I assume that the statement we want to prove is lim_(xrarr-3)(x^2+3x)=0
Preliminary Work
The epsilon, delta proofs for quadratic functions are more involved than those for linear functions. Most of us need to be shown this method, we don't come up with it by ourselves.
We begin (as we do for linear functions) by examining the difference between f(x) and L.
Keep in mind that, for his question #abs(x-a) = abs(x-(-3)) = abs(x+3)
Now,
abs((x^2+3x)-0) = abs(x(x+3)) = absx abs(x+3)
We can control, through the choice of delta, the size of abs(x+3).
If abs(x+3) is small, then x is close to -3. We'll put a bound on absx by choosing delta so that is is less than some positive number we select now.
1 is easy to work with, so we'll make sure that delta <= 1.
This will assure us that
when abs(x-(-3)) < delta <= 1, we have
x is within 1 unit of -3,
So, -3-1 < x < -3+1
I.e. -4 < x , -2,
so absx < 4.
If we also make sure that delta <= epsilon/4, the we will have:
When abs(x+3) < delta, then
absx abs(x+3) < 4abs(x+3) < 4(epsilon/4) = epsilon
Proof
Given epsilon > 0, choose delta = min{1, epsilon/4}
If 0 < abs(x+3) < delta, then
abs((x^2+3x)-0) = absx abs(x+3) < 4abs(x+3) <= 4(epsilon/4)=epsilon
That is, abs((x^2+3x)-0) < epsilon.
