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#