How to prove $\lim_{x\rightarrow -1^+}\sqrt{x+1}=0$ using $\epsilon-\delta$ definition? Use the $\epsilon-\delta$ definition to prove:
$$\lim_{x\rightarrow -1^+}\sqrt{x+1}=0$$
I don't understand what to do with: $$x\rightarrow -1^+$$
 A: In a normal two-sided limit, we'd want to prove that for all $\epsilon > 0$, there is some $\delta > 0$ such that:
$$
0 < |x + 1| < \delta \implies |\sqrt{x + 1} - 0| < \epsilon
$$
But since we are approaching $-1$ from the right, we are assuming that $x > -1$, which implies that $x + 1 > 0$ (so that we aren't taking the square root of a negative number). This allows us to safely drop the absolute values, so we may simplify the above implication a bit to become:
$$
0 < x + 1 < \delta \implies |\sqrt{x + 1} - 0| < \epsilon
$$

Indeed, given any $\epsilon > 0$, let $\delta = \epsilon^2 > 0$. Then if $0 < x + 1 < \delta$, notice that:
\begin{align*}
|\sqrt{x + 1} - 0|
&= \sqrt{x + 1} \\
&< \sqrt{\epsilon^2} &\text{since }x + 1 < \delta = \epsilon^2\\
&= \epsilon
\end{align*}
as desired. $~~\blacksquare$
A: The notation used means that $x$ goes to $-1$ from above (else the function wouldn't be real-valued), alternative notation is
$$\lim_{x\searrow -1} \sqrt{1+x} = \lim_{x\to-1, x>-1} \sqrt{1+x} = \lim_{x\to-1^+} \sqrt{1+x}$$
So you have to show that $\forall\epsilon > 0\ \exists\delta > 0$ such that
$$\sqrt{1+(-1+h)} <\epsilon \quad\forall 0< h<\delta$$
Where normal $\epsilon-\delta$ (for the two-sided limit) would require $\forall h: |h|<\delta$.
