#lim_(x->-2) x^2 = 4#
Evaluate the expression:
#abs(x^2-4) = abs((x+2)(x-2))= abs(x-2)abs(x+2)#
if we write #x = xi +2#, then:
#abs(x^2-4) = abs(xi)abs(xi+4)#
and using the triangular inequality:
#abs(x^2-4) <= abs(xi)(abs(xi)+4)#
#abs(x^2-4) <= abs(xi)^2 +4 abs(xi)#
Now consider the trinomial:
#q^2+4q-1/2 #
as this is a trinomial with positive leading coefficient we know that its value is negative for values of #q# in the interval between the two roots:
#q= (-2+-sqrt(4+1/2)) = -2+-3/sqrt2 = (+-3sqrt2 -4)/2#
so if we take:
#delta_epsilon <= (3sqrt2 -4)/2#
we have:
#delta_epsilon^2+4delta_epsilon -1/2 < 0#
which means:
#delta_epsilon^2+4delta_epsilon < epsilon#
and then for #x in (-2-delta_epsilon, 2+delta_epsilon)# and consequently #abs (xi) < delta_epsilon#:
#abs(x^2-4) <= abs(xi)^2 +4 abs(xi) < delta_epsilon^2+4delta_epsilon < epsilon#
which proves the point.
