How do you prove that the limit of (x^2 - 4x + 5) = 1(x24x+5)=1 as x approaches 2 using the epsilon delta proof?

1 Answer
May 10, 2017

See below.

Explanation:

Problem:
lim_(x->2)(x^2-4x+5)=1

Work:
|(x^2-4x+5)-1|< epsilon
|x^2-4x+4|< epsilon
|(x-2)|^2< epsilon
|x-2|< sqrt(epsilon)

Proof:

AA epsilon>0, EE delta>0 such that
if 0<|x-2|< delta, then |(x^2-4x+5)-1|< epsilon
Given 0<|x-2|< delta, let delta=sqrt(epsilon)
|x-2|< sqrt(epsilon)
|x-2|^2< epsilon
|(x-2)^2|< epsilon
|x^2-4x+4|< epsilon
|(x^2-4x+5)-1|< epsilon
thereforelim_(x->2)(x^2-4x+5)=1