Question

How do you use the epsilon delta definition to prove that the limit of `x^3=27` as `x->3`?

Answer 1

Given `epsilon > 0`. let `delta = min{1,epsilon/37}`

Whenever `0 < abs(x-3) < delta`

then `abs(x^2+3x+9) < 37`, so

`abs(x^3-27) = abs(x-3)abs(x^2+3x+9)`

`< abs(x-3) * 37`

`< epsilon/37 * 37 = epsilon`

That is, if `0 < abs(x-3) < delta`, then `abs(x^3-27) < epsilon`