Question #4c5ac

2 Answers
Jun 20, 2017

See below.

Explanation:

Prove: #lim_(x->-3)x^2+4x+1=-2#

Work (not part of proof):

#0 < |x+3| < delta#, #|(x^2+4x+1)+2| < epsilon#

We need to manipulate the #|(x^2+4x+1)+2| < epsilon# to be #|x+3|<"something"# to set delta equal to that term:
#|(x^2+4x+1)+2| < epsilon#

#|x^2+4x+3| < epsilon#

#|(x+3)(x+1)| < epsilon#

#|x+3| < epsilon/|x+1|#

Since we cannot have an #x# term with epsilon, we let #delta=1# and solve for the value of #x+1#:
#0 < |x+3| < 1#

#-1 < x+3 < 1#

#-3 < x+1 < -1#

Here, we choose the larger value since if we chose the smaller value, the #-3# would not be included, so:
#|x+1|<3#

Therefore,
#|x+3| < epsilon/3#

Proof:

#forall# #epsilon > 0#, #exists# #delta > 0# such that:
if #0 < |x+3| < delta#, then #|(x^2+4x+1)+2| < epsilon#.
Given #0 < |x+3| < delta#, let #epsilon = min(1,epsilon/3)#:

#0 < |x+3| < epsilon/3#

#0 < 3|x+3| < epsilon#

#0 < |x+1||x+3| < epsilon#

#0 < |x^2+4x+3| < epsilon#

#0 < |(x^2+4x+1)+2| < epsilon#

#therefore# #lim_(x->-3)x^2+4x+1=-2#

Jun 20, 2017

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.