Given lim as x-->-2 of 2x-2 is -8, what is the best choice of d such that |(2x-2)-(-8)| < 0.03 whenever |x+2|<d?
--> I tried to solve |(2x-2)-(-8)| to get it in terms of |x+2| but I can't seem to get it.
I had |2x+6| = 2|x+3| but that doesn't help much.
Also:
b) When the value of x is within 1/20 of 1, how close is the value of f(x) = 4x+5 to 9?
--> I have:
0 < |x-1| < 0.05
|(4x+5)-9| < E
But now I'm stuck....