Since f(x) = x^2+4x+1 is a polynomial, it is continuous for every x in RR, then:
lim_(x->-3) f(x) = (-3)^2+4*(-3)+1 = 9-12+1 = -2
We want to prove that given epsilon >0 we can find a corresponding delta_epsilon such that:
abs(x-(-3)) < delta_epsilon => abs (x^2+4x+1-(-2)) < epsilon
that is:
abs(x+3) < delta_epsilon => abs (x^2+4x+3) < epsilon
Evaluate:
abs (x^2+4x+3) = abs((x+3)(x+1)) = abs(x+3)abs(x+1)
Now for abs(x+3) < delta_epsilon we have:
abs(x+1) = abs(x+3-2)
and based on the triangular inequality:
abs(x+1) <= abs(x+3)+abs(2)
that is:
abs(x+1) < delta_epsilon+2
Given an arbitrary number epsilon > 0 we can choose then delta_epsilon < min(1, epsilon/3) and we have that because delta_epsilon < 1:
abs (x+1) < 2+delta_epsilon < 2+1 = 3
and because delta_epsilon < epsilon/3
abs(x+3) < delta_epsilon < epsilon/3
but then:
abs(x+3)abs(x+1) < 3 xx epsilon/3 = epsilon
and in conclusion:
abs (x^2+4x+3) < epsilon
which proves the point.