to make that less than , you need to have . But we need a constant, not a function of x on the right. That is, we need which means we need an lower bound on . If |x-1|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as |x-1|< 1, |x+1|> 2 so [tex]\frac{1}{|x+1|}< \frac{1}{2}. Then . In order that both of those be true, you must have |x-1|< the smaller of 1 and . Assuming that n is a positive integer, of course, is less than 1 and so is the smaller. The condition you want is .