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

Question

Ezra Adams · Accepted Answer

See below.

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#

Emilia Aguilar · Answer

undefined To prove that the limit of (x^2 - 4x + 5) as x approaches 2 is equal to 1 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that whenever 0 < |x - 2| < delta, then |(x^2 - 4x + 5) - 1| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that |(x^2 - 4x + 5) - 1| < epsilon whenever 0 < |x - 2| < delta. First, let's simplify the expression |(x^2 - 4x + 5) - 1|: |(x^2 - 4x + 4)| = |(x - 2)(x - 2)| = |x - 2|^2 Now, we want to find a delta such that |x - 2|^2 < epsilon whenever 0 < |x - 2| < delta. Taking the square root of both sides, we have: |x - 2| < sqrt(epsilon) So, if we choose delta = sqrt(epsilon), then whenever 0 < |x - 2| < delta, we have |(x^2 - 4x + 5) - 1| < epsilon. Therefore, we have proven that the limit of (x^2 - 4x + 5) as x approaches 2 is equal to 1 using the epsilon-delta proof.