I have asked the question about the limit of two functions as they're approaching the real axis before. Today I considered the function $f(z)=\sqrt{1-z^2}$, trying to calculate the difference in similar way.
The function have two branch points $z=\pm1$. When it takes value above the line segment $(-1,1)$, I split it into two parts $-1<x<0$ and $0<x<1$. I chose the principal value of $f(z)$ and let $-\pi<\text{arg}(z)\leq\pi$. \begin{equation} f(z)=\sqrt{1-z^2}=e^{\frac{1}{2}\log(1-z^2)}=e^{\frac{1}{2}(\log(\mid 1-z^2\mid)+i\text{arg}(1-z^2))} \end{equation} Substitute $z=x+iy$, then $f(x+iy)=\sqrt{\mid1-x^2+y^2\mid}e^{\frac{i}{2}\text{arg}(1-x^2+y^2-2ixy)}$. When $y\to0+$, the magnitude tends to $\sqrt{1-x^2},x\in(-1,1)$, and the argument becomes \begin{equation} \text{arg}(1-x^2+y^2-2ixy)= \begin{cases} 0-, & 0<x<1\\ 0+, & -1<x<0 \end{cases} \end{equation} So the limit of $f(x+iy)$ as $z=x+iy$ approaching the line segment $(-1,1)$ is always $\sqrt{1-x^2}$. Do another similar operation to $f(x-iy)$ as $y\to0+$, $f(z)=\sqrt{\mid 1-x^2+y^2\mid}e^{\frac{i}{2}\text{arg}(1-x^2+y^2+2ixy)}$, we have \begin{equation} \text{arg}(1-x^2+y^2+2ixy)= \begin{cases} 0+, & 0<x<1\\ 0-, & -1<x<0 \end{cases} \end{equation} It comes out the same result as previous one. So the limit \begin{equation} \lim_{y\to0+}(f(x+iy)-f(x-iy))=0,\ -1<x<1 \end{equation} exists and is identically zero. Is this solution feasible? Can we have better methods?
P.S.: The visualization shows the same result.
P.P.S.: I also calculated the situation when $x>1$ and $x<-1$, the difference when $y\to0+$ should be \begin{equation} \begin{cases} -2i\sqrt{\mid 1-x^2\mid}, & x>1\\ 2i\sqrt{\mid 1-x^2\mid}, &x<-1 \end{cases} \end{equation}