use epsilon delta to prove the following
lim x^2-9/x+3 =-6
x->-3
f(x) = (x^2-9)/(x+3) = (x-3)
Obviously the function is -6 when x = -3. You don't need to prove it with epsilons.
If you want to use epsilon, which I will call e, let x = -3 + e
f(x) = (9 -6e +e^2 -9)/e
= -6 +e
= -6 as e-> 0