I am looking at $f(z)=\sqrt{1-z^2}$ and a branch cut on the real axis from $z=-1$ to $z=1$.
Is it correct to say that $f(x+i\epsilon) = -f(x-i\epsilon)$ when $x\in(-1,1)$, $\epsilon\in\mathbf{R}$ and $\epsilon\to 0$? If this is true, is it also true that $f(x+iy)=-f(x-iy)$ when $y>0$?
(The problem I'm looking at is similar to Dustins answer to Integral with branch cut ( Problem while calculating residue) and I (think) understand why integration over and under the branch cut doesn't eliminate each other. What I don't understand is the minus sign when he is summing the residues, and my concern is if it has anything to do with my question above.)
Write $f$ in the form $$f(z):={z\over i}\>{\rm pv}\sqrt{1-{1\over z^2}}\qquad\bigl(z\in\Omega:={\mathbb C}\setminus[{-1},1]\bigr)\ .$$
It is easily checked that for $z\in\Omega$ the expression $1-{1\over z^2}$ cannot be real $\leq0$. It follows that for $z\in\Omega$ the principal value ${\rm pv}\sqrt{1-{1\over z^2}}$ and therewith $f$ is well defined and analytic in $\Omega$.
The principal value $w\mapsto{\rm pv}\sqrt{w}$ has the symmetry property $${\rm pv}\sqrt{\bar w}=\overline{{\rm pv}\sqrt{w}}$$ on its domain of definition. It follows that $f$ has the symmetry properties $$f(\bar z)={\bar z\over i}{\rm pv}\sqrt{1-{1\over \bar z^2}}=-\overline{f(z)}\ ,\qquad f(-z)=-f(z)\qquad(z\in\Omega)\ .\tag{1}$$ Writing $z=x+iy$ one has $$f(x-iy)=-\overline{f(x+iy)}\ ,$$ and this shows that the second of your conjectures only is valid at the points $z\in\Omega$ where $f$ happens to asssume a real value. (These are the points $\ne0$ on the imaginary axis.)
In order to discuss the $\lim f(x+iy)$ for $y\to0+$ and $y\to0-$ when $|x|<1$ we shall argue as follows: The value $f(i)=\sqrt{2}$ coincides with the principal value of the originally given expression $\sqrt{1-z^2}$ at $i$. Furthermore $1-z^2=1-x^2+y^2-2ixy$ is not real $\leq0$ in the open upper halfplane $H_+$: It can only be real on $H_+$ when $x=0$, and in this case $1-z^2=1+y^2>0$. It follows that in fact $$f(z)={\rm pv}\sqrt{1-z^2}={\rm pv}\sqrt{1-x^2+y^2-2ixy}\qquad(z\in\Omega\cap H_+)\ .$$ This immediately implies $$\lim_{y\to0+} f(x+iy)=\sqrt{1-x^2}\qquad\bigl(|x|<1)\ .$$ From the second identity $(1)$ we then infer that $$f(z)=-{\rm pv}\sqrt{1-z^2}\qquad(z\in\Omega\cap H_-)\ ,$$ and in particular $$\lim_{y\to0-} f(x+iy)=-\sqrt{1-x^2}\qquad\bigl(|x|<1)\ .$$ Therefore the first of your conjectures is correct.