$\varepsilon$-$\delta$ 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?
 A: If you want to show that $\lim_{x \to 0} \sqrt{1+x} = 1$, you need to: take $\epsilon > 0$ arbitrary and find a $\delta > 0$ such that if $|x|<\delta$ then $|\sqrt{1+x} - 1| < \epsilon$. 
Hint: The difference of squares formula gives $$ x = (1+x)-1 = [\sqrt{1+x}+1][\sqrt{1+x}-1]$$ so $$ \big| \sqrt{1+x} - 1 \big| \leq \left| \frac{x}{\sqrt{1+x}+1}\right| = \frac{|x|}{\sqrt{1+x}+1}$$ Now, can you continue from here since you are allowed to choose $\delta$ and note $|x|<\delta$?
A: 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.
A: 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 $!
