Let #x=1+xi# 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)#

Given any number #epsilon > 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#