in fact, $\displaystyle \delta = \frac {\epsilon}2$ would work i think
anyway, recall what the definition of a limit means.
we want to find a $\displaystyle \delta > 0$, such that for all $\displaystyle \epsilon > 0$ (and $\displaystyle x \in \text{dom}(f)$), $\displaystyle |x - 1| < \delta$ implies $\displaystyle |f(x) - 2| < \epsilon$.
now, we found that $\displaystyle |x - 1| < \frac {\epsilon}{x + 1}$, but we want $\displaystyle |x - 1|< \delta$, so we need to find a $\displaystyle \delta$ that works. now, what does it mean for $\displaystyle x$ to be close to 1? we give ourselves a generous range. and say, let it be somewhere between 0 and 2, and those x's are "close". now, to make sure our $\displaystyle \delta$ works, we decide to be cautious and choose $\displaystyle x = 2$, that way, $\displaystyle x$ gets really close to 1, since $\displaystyle \frac {\epsilon}{x + 1}$ is the smallest for the range we are considering. so, if $\displaystyle x = 2$, we have $\displaystyle \frac {\epsilon}{x + 1} = \frac {\epsilon}3$, and so we choose that as our $\displaystyle \delta$.
ok, so that was hopelessly confusing, even to me, i know what i want to say, but i am not sure if i said it ok. did you get that?
actually, it is delta that depends on epsilon, not the other way around. we must choose a delta that works for any epsilon we are given