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.
