Verified answer

You seem to be missing some parentheses:

1/(x+1), right? If not, please speak up, in Additional Details.

The exercise of "epsilondics" isn't very suitable for *finding* a limit value; it's more about proving it rigorously once you've found it. In this case, knowing (or suspecting) that

f(x) = 1/(x+1)

is a continuous function for x ≠ -1, you just substitute x = 0:

f(0) = 1/(1+0) = 1

lim[x→a] f(x) = b ⇔ ( ∀ε>0 ∃δ ∍ |x-a|<δ ⇒ |f(x)-b|<ε )

And for this case, we have a=0, b=1.

So the game is, given ε, to find δ, in order to show that it satisfies the "such that"

(∍ |x-a|<δ ⇒ |f(x)-b|<ε), which, here, is

(∍ |x|<δ ⇒ |f(x)-1|<ε)

|f(x) - b| = |1/(x+1) - 1| = |x/(x+1)| = |x|/|x+1| < δ/|x+1|

and since |x|<δ, thus -δ<x<δ, so if 0<δ<1, |x+1| ≤ 1-δ, so

δ/|x+1| ≤ δ/(1-δ) = 1/(1-δ) - 1 = ε

δ = ε/(1+ε) = 1/(1/ε + 1)

So if ε = 1/N, δ = 1/(N+1)

This marks success, because for any positive ε, no matter how small, we can produce a δ that satisfies the definition for the limit.

PS. This process is always very "thought-dense," so you'll probably need a lot of concentration to follow it all the way through. If you get stuck, pipe up! If you can understand it all in the first pass, you're better at it than I am!