Question

How do you prove that the limit of `x^2 - 5x + 6 = 2` as x approaches 1 using the epsilon delta proof?

Answer 1

See below.

Explanation:

Given any number `epsilon > 0` choose `delta = min (1, epsilon/4)` then we have that:

For any `x` with `0 < abs(x-1) < delta`,

`-1 < x-1 < 1`, so `-4 < x-4 < -2`

thus `abs(x-4) < 4`

and

`abs(x-1) < epsilon/4`

so that:

If `0 < abs(x-1) < delta` then

`abs((x^2-5x+6) -2) = abs(x-4)abs(x-1) < 4abs(x-1) < 4delta = 4 epsilon/4 = epsilon`