We have to prove that:
#lim_(x->3) f(x) = f(3)#
that is:
#lim_(x->3) x^2= 9#
Evaluate the difference:
#abs(x^2-9) = abs((x-3)(x+3)) = abs(x-3)abs(x+3)#
and posing:
#xi = x-3#
#abs(x^2-9) = absxi (abs(xi+6))#
Using the triangular inequality:
#abs(x^2-9) <= absxi (abs(xi)+6)#
#abs(x^2-9) <= absxi^2 +6 abs(xi)#
Given any #epsilon >0# choose now #delta_epsilon < min(1,epsilon/7)#
For #x in (3-delta_epsilon, 3+delta_epsilon)# we have that #abs(xi) < delta_epsilon#.
Now, as #absxi < delta_epsilon < 1# we have that #absxi^2 < abs xi#, then:
#abs(x^2-9) <= absxi^2 +6 abs(xi) < 7abs(xi)#
and as: #absxi < delta_epsilon < epsilon/7#
#abs(x^2-9) < 7abs(xi) < 7 epsilon /7 = epsilon#
In conclusion, for any #epsilon > 0# by choosing #delta_epsilon < min(1,epsilon/7)# we have that:
#x in (3-delta_epsilon, 3+delta_epsilon) => abs(x^2-9) < epsilon#
which proves the point.
