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+1−3∣∣=∣∣2(1+ξ)2−2∣∣=∣∣2+4ξ+2ξ2−2∣∣=∣∣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