Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist. Show that the limit $\lim_{z \to i}{\frac{1-|z|}{i-z}}$ does not exist.
I tried by putting $z =ix $, where $x \to 1$ and got that the limit is $\frac{1}{i}$, but can't think of another example. 
 A: Hint:
$$\lim_{z \to i}{\frac{1-|z|}{i-z}}= \lim_{s \to 0}{\frac{|s+i|-1}{s}}.$$
Choose $s$ real number then the limit will be 0.
A: Choose some $k\in\mathbb{C}$ and $\epsilon\in\mathbb{R}$ and substitute $z=i+\epsilon k$ then we have the limit equal to
$$\begin{align}
\lim_{\epsilon\to0}\frac{1-|i+\epsilon k|}{i-(i+\epsilon k)}
&=\lim_{\epsilon\to0}\frac{1-\sqrt{(\Re{(\epsilon k)})^2+(1+\Im{(\epsilon k)})^2}}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{1-\sqrt{1+(\Re{(\epsilon k)})^2+2\Im{(\epsilon k)}+(\Im{(\epsilon k)})^2}}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{1-(1+\frac12((\Re{(\epsilon k)})^2+2\Im{(\epsilon k)}+(\Im{(\epsilon k)})^2+O(\epsilon^3))}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12(\Re{(\epsilon k)})^2-\Im{(\epsilon k)}-\frac12(\Im{(\epsilon k)})^2+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12(\epsilon\Re{(k)})^2-\epsilon\Im{(k)}-\frac12(\epsilon\Im{(k)})^2+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac{-\frac12\epsilon^2\Re^2{(k)}-\epsilon\Im{(k)}-\frac12\epsilon^2\Im^2{(k)}+O(\epsilon^3)}{-\epsilon k}\\
&=\lim_{\epsilon\to0}\frac1k\left(\frac12\epsilon\Re^2{(k)}+\Im{(k)}+\frac12\epsilon\Im^2{(k)}+O(\epsilon^2)\right)\\
&=\frac1k\Im{(k)}\\
\end{align}$$
Where $\frac1k\Im{(k)}$ is an arbitrary complex number, so the limit does not exist.
