Question

How do you prove that the limit of `(4 + x - 3x^3) = 2` as x approaches 1 using the epsilon delta proof?

Answer 1

See below.

Explanation:

Note that `(4+x-3x^3)-2 = -(3x^3-x-2) = -(x-1)(3x^2+3x+2)`

So `abs((4+x-3x^3)-2) = abs(x-1) * abs(3x^2+3x+2)`

Given `epsilon > 0`, let `delta = min{1, epsilon/20}` and note that `delta > 0`

If `0 < abs(x-1) < 1`
then `0 < x < 2` so that `2 < (3x^2+3x+2) < 20` and

`abs(3x^2+3x+2) < 20`

So, `abs((4+x-3x^3)-2) = abs(x-1) * abs(3x^2+3x+2)`

`< abs(x-1) * 20 < epsilon/20 * 20 = epsilon`

We have shown that for any positive `epsilon`, there is a positive `delta` such that

If `0 < abs(x-1) < 1` , then `abs((4+x-3x^3)-2) < epsilon`

So, by the definition of limit, `lim_(xrarr1)(4+x-3x^2) = 2`