$\varepsilon$-$\delta$ proof of $\sqrt{x+1}$ need to prove that $ \lim_{x\rightarrow 0 } \sqrt{1+x} = 1 $
proof of that is:
need to find a delta such that $ 0 < |x-1| < \delta \Rightarrow 1-\epsilon < \sqrt{x+1} < \epsilon + 1 $ if we choose $ \delta = (\epsilon + 1)^2 -2 $ and consider $ |x-1| < \delta = (\epsilon + 1)^2 - 2 $
$ 4 - (\epsilon + 1)^2 < x +1 < (\epsilon + 1)^2 $
$ \sqrt{4-(\epsilon + 1)^2} -1 < \sqrt{x+1} -1 < \epsilon$ but I need to show that $\sqrt{4-(\epsilon + 1)^2} -1 > -\epsilon $ before this proof is complete...
any help on how to finish the proof?

So it remains to show that $\sqrt{4-(\epsilon + 1)^2} -1 > -\epsilon $.
Note that for the $\epsilon-\delta$ proof is enough to show that $|x-1|\lt\delta\Rightarrow\ldots$ only for small epsilons. In your case we may assume therefore that $\epsilon<\frac12$. In that case we have that indeed $\sqrt{4-(\epsilon + 1)^2} -1 > -\epsilon $!
