Originally Posted by
HallsofIvy $\displaystyle |x^2- 1|= |x-1||x+1|$ to make that less than $\displaystyle \frac{1}{n}$, you need to have $\displaystyle |x- 1|< \frac{1}{n|x+1|}$. But we need a constant, not a function of x on the right. That is, we need $\displaystyle |x- 1|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\displaystyle \frac{1}{x+1}$. 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 $\displaystyle |x-1|< \frac{1}{n|x+1|}< \frac{1}{2n}$. In order that both of those be true, you must have |x-1|< the smaller of 1 and $\displaystyle \frac{1}{2n}$. Assuming that n is a positive integer, of course, $\displaystyle \frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\displaystyle |x-1|< \frac{1}{2n}$.