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.