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.