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
