How do you prove that the limit of (2x^2 + 1) = 3 (2x2+1)=3 as x approaches 1 using the epsilon delta proof?

1 Answer
Jun 7, 2018

Let x=1+xix=1+ξ so that:

abs(2x^2+1-3) = abs(2(1+xi)^2-2) = abs (2+4xi+2xi^2-2) = abs(4xi+2xi^2) = 2abs(xi)abs(2+xi)2x2+13=2(1+ξ)22=2+4ξ+2ξ22=4ξ+2ξ2=2|ξ||2+ξ|

Given any number epsilon > 0ε>0 choose delta_epsilon < min(1, epsilon/6).

Then as delta_epsilon < 1 we have that x in (1-delta_epsilon, 1+delta_epsilon) implies that abs xi < 1 and:

abs (2+xi) <= 2 +abs xi < 3

Furthermore, as delta_epsilon < epsilon/6 we have that x in (1-delta_epsilon, 1+delta_epsilon) implies abs xi < epsilon/6 and then:

abs(2x^2+1-3) = 2abs(xi)abs(2+xi) < 2*epsilon/6 *3 = epsilon

Thus given epsilon > 0 and delta_epsilon < min(1,epsilon/6):

x in (1-delta_epsilon, 1+delta_epsilon) => abs(2x^2+1-3) < epsilon

which proves that:

lim_(x->1) 2x^2+1 = 3