$\varepsilon$-$\delta$ 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?

You need to show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon.$$
It's helpful in this case to multiply by the conjugate of $\sqrt{x+1}-1$, which is $\sqrt{x+1}+1$, so that the rightmost inequality becomes $$|\sqrt{x+1}-1|\cdot|\sqrt{x+1}+1|<\varepsilon(\sqrt{x+1} + 1)$$ $$|x|<\varepsilon(\sqrt{x+1}+1). \tag{1}$$
Now, if you require that $0 < |x| < 1$, then we know that $0<\sqrt{x+1}<\sqrt{2}$, so $1<\sqrt{x+1} + 1<\sqrt{2}+1.$
So we now know that if we require $0 < |x| < 1$, then $(1)$ tells us that $|x| < \varepsilon(\sqrt{2}+1)$,  which we have also shown can be rearranged into the form $|\sqrt{x+1}-1| < \varepsilon.$
So, we take $\delta = \min\left(1,\ \varepsilon(\sqrt{2}+1)\ \right).$ Then, $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon$$
which is exactly what we wanted to show.
