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since this is a line; your delta is simple. :)

lim{x->6} (2x+1) = 13
f(d) = 2d+1 = 13+e
2d = 12+e
d = (12+e)/2 i think

maybe.... es and ds tend to confuse me lol

or should that be f(x+d)??

d < e/2 is good

i can see it, but trying to describe it eludes me....

|2x+1-13|=|2x-12|=2*|x-6|