Question

How do you use the epsilon delta definition to prove that the limit of `x^3+6x^2=32` as `x->2`?

Answer 1

`delta(epsilon)=min(1;epsilon/49)`

Explanation:

`abs(x^3+6x^2-32)=abs(x-2)(x+4)^2`
`abs(x-2) < min(1;epsilon/49) rArr (x+4)^2<(2+1+4)^2=49`
and
`abs(x^3+6x^2-32) < epsilon/49 * 49 = epsilon`