How do you prove that the limit of #(x^2 - 1) = 3# as x approaches -2 using the epsilon delta proof?

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

Work (not part of proof): #0<|x+2|< delta#; #|(x^2-1)-3|< epsilon#

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

#|(x^2-1)-3|< epsilon#

#|x^2-4|< epsilon#

#|(x+2)(x-2)| < epsilon#

#|x+2| < epsilon/(x-2)#

Since we cannot have a #x# term with epsilon, we let #delta = 1# and solve for the value #x+2# would be: #0 < |x+2| < 1#

#-1 < x+2 < 1#

#-1-4 < x+2-4 < 1-4#

# -5 < x-2 < -3 #

Here, we choose the larger value since if we chose the smaller value, the -5 would not be included, so: #|x-2|<5#

Therefore, #|x+2|< epsilon/5#

Proof: #forall# #epsilon>0#, #exists# #delta>0# such that: if #0<|x+2|< delta#, then #|(x^2-1)-3|< epsilon#. Given #0<|x+2|< delta#, let #delta = min(1,epsilon/5)#: #0<|x+2|< epsilon/5#

#0<5|x+2|< epsilon#

#0<|x-2||x+2|< epsilon#

#0<|x^2-4|< epsilon#

#0<|(x^2-1)-3|< epsilon#

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

To prove that the limit of (x^2 - 1) is 3 as x approaches -2 using the epsilon-delta proof, we need to show that for any given epsilon greater than 0, there exists a corresponding delta greater than 0 such that whenever 0 < |x - (-2)| < delta, then |(x^2 - 1) - 3| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a delta > 0 such that whenever 0 < |x - (-2)| < delta, then |(x^2 - 1) - 3| < epsilon. First, let's simplify the expression |(x^2 - 1) - 3|: |(x^2 - 1) - 3| = |x^2 - 4| Now, we can factor the expression x^2 - 4 as (x - 2)(x + 2): |x^2 - 4| = |(x - 2)(x + 2)| Since we are interested in values of x approaching -2, we can assume that x is close to -2, but not equal to -2. Therefore, we can assume that |x + 2| > 0. Now, let's choose a value for delta. We want to find a delta such that whenever 0 < |x - (-2)| < delta, then |(x^2 - 1) - 3| < epsilon. Let's consider the inequality |(x - 2)(x + 2)| < epsilon. Since |x + 2| > 0, we can divide both sides of the inequality by |x + 2|: |(x - 2)(x + 2)| / |x + 2| < epsilon / |x + 2| Simplifying further, we have: |x - 2| < epsilon / |x + 2| Now, we can choose a value for delta. Let's set delta = min(1, epsilon / 5). If 0 < |x - (-2)| < delta, then we have: |x - (-2)| < delta Simplifying, we get: |x + 2| < delta Since delta = min(1, epsilon / 5), we have: |x + 2| < min(1, epsilon / 5) Now, let's consider the inequality |x - 2| < epsilon / |x + 2|. Since |x + 2| < min(1, epsilon / 5), we can substitute this value: |x - 2| < epsilon / |x + 2| < epsilon / min(1, epsilon / 5) Since epsilon > 0, we can assume that epsilon / min(1, epsilon / 5) > 0. Therefore, we can choose delta = min(1, epsilon / 5) as our delta value. To summarize, for any given epsilon > 0, we have shown that there exists a corresponding delta = min(1, epsilon / 5) > 0 such that whenever 0 < |x - (-2)| < delta, then |(x^2 - 1) - 3| < epsilon. This proves that the limit of (x^2 - 1) is 3 as x approaches -2 using the epsilon-delta proof.