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

1 Answer
May 2, 2017

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