How do you prove that the limit #((x^2+x-6)/(x-2))=5# as x approaches 2 using the formal definition of a limit?

Question

Elijah Abram · Accepted Answer

below

#lim_{x to 2} ((x^2+x-6)/(x-2))# let #x = 2+h# #= lim_{h to 0} (((2+h)^2+(2+h)-6)/((2+h)-2))# #= lim_{h to 0} (4+4h+ h^2+2+h-6)/(h)# #= lim_{h to 0} (5h+ h^2)/(h)# #= lim_{h to 0} 5 + h# #=5#

Scarlett Allison · Answer

undefined To prove that the limit of ((x^2+x-6)/(x-2)) as x approaches 2 is equal to 5 using the formal definition of a limit, we need to show that for any given positive value ε, there exists a positive value δ such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε. Let's proceed with the proof: Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε. First, let's simplify the expression ((x^2+x-6)/(x-2)) - 5: ((x^2+x-6)/(x-2)) - 5 = ((x^2+x-6) - 5(x-2))/(x-2) = (x^2+x-6 - 5x + 10)/(x-2) = (x^2 - 4x + 4)/(x-2) = (x-2)(x-2)/(x-2) = x-2. Now, we want to find a δ > 0 such that if 0 < |x - 2| < δ, then |x - 2| < ε. Since we want to prove that the limit is 5, we can choose δ = ε. If 0 < |x - 2| < δ = ε, then |x - 2| < ε. Therefore, we have shown that for any given positive value ε, there exists a positive value δ (in this case, δ = ε) such that if 0 < |x - 2| < δ, then |((x^2+x-6)/(x-2)) - 5| < ε. Hence, using the formal definition of a limit, we have proven that the limit of ((x^2+x-6)/(x-2)) as x approaches 2 is equal to 5.