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#
