Inversion in a circle as radius goes to infinity. I am trying to show that the in the limit case as the circle gets very large, inversion in it is equivalent to reflection in a line. I have the transform $z \to c+ \frac{R^2}{ (\overline z -\overline c)} $ for inversion of $z$ in a circle radius $R$ centred at $c$. I am unsure how to go about taking a limit to get the result as limit as R goes to infinity does not too exist. As such I assume you need to take a limit less than infinity. How would you go about finding a suitable limit case?  
 A: You are right that the statement you want is false without more care taken in the formulation of its statement.
In order to get a true statement, you need to let the center of the circle vary as its radius varies.
Here's a true statement. For each value of the radius $R > 0$ let $P_R$ denote the point $R+0i$ in the complex plane, and let $C_R$ denote the circle of radius $R$ centered on $P_R$. Note that $C_R$ is tangent to the imaginary axis at the origin. The true statement is that as we let $R$ approach $+\infty$, inversion in the circle $C_R$ converges to reflection across the imaginary axis.
A: Fix a point on the circle and send the center to infinity. 
A: For clarity, given a circle $C$ of radius $R$ centred at some point $c\in\mathbb{C}$, let $t$ be the translation sending $c$ to the origin, $d$ the dilation of the plane by $1/R$, and $I$ inversion in the unit circle, then inversion $I_C$ in $C$ is equal to the product $t^{-1}d^{-1}Idt$. Thus $$\tag{*}I_C(z)=\frac{R^2}{\left(\overline{z}-\overline{c}\right)}+c.$$ We wish to prove that reflection in a line is a limiting case of inversion in a circle. This means, given any line $\mathscr{L}$ tangent to $C$, if we keep the point of intersection fixed and let $R\rightarrow\infty$ by moving $c$, then $I_C$ tends to reflection $r_{\small{\mathscr{L}}}$ in $\mathscr{L}$.
Since rotations and translations are isometries, it suffices to prove the result for reflection $r$ in the line $1+i\mathbb{R}$ (or the imaginary axis $i\mathbb{R}$, or any vertical line; I just like to start with the unit circle): $$r(z)=2-\overline{z}.$$ Further, we can take $z$ to be any point on the $x$-axis other than $1$, so that $r(z)=2-z$.
Hence, for given $\delta\geq 0$ we consider a circle of radius $R=1+\delta$ centred at $c=-\delta$ on the real line. Plugging these values into $(*)$ we get $$\frac{(1+\delta)^2}{z+\delta}-\delta=\frac{1+\delta(2-z)}{z+\delta}\rightarrow 2-z\space\text{as}\space \delta \rightarrow\infty.$$
