Here is a major hint: **log** in some contexts may actually mean **ln** when the variables is real. Also, note that the absolute value **|.|** maps from $\mathbb{C}\mapsto \mathbb{R}^+$. Hence, **log|z-2|** is just a real logarithm, and is in fact defined everywhere on the complex plane except for z=2. This can be done without using the delta-epsilon definition of limits (which is unfortunately hard to verify).

We have the expression $\lim\limits_{z \to 2} (z-2) \ln|z-2|$. The lim is of the form $0\cdot-\infty$, meaning L'Hôpital's Rule applies.

Note that for L'Hôpital's Rule, $0 = \frac{1}{\infty}$. Hence, $(z-2) = \frac{1}{(z-2)^{-1}}$, and the lim of both when they approach 2 are zero.

$\lim\limits_{z \to 2} (z-2) \ln|z-2| = \lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}}$, which is of the $\frac{\infty}{\infty}$ form, the desired indeterminate form.

We apply L'Hôpital's Rule: $\lim\limits_{z \to 2}\frac{\ln|z-2|}{(z-2)^{-1}} \Rightarrow \lim\limits_{z \to 2}\frac{(z-2)^{-1}}{-1(z-2)^{-2}}=\lim\limits_{z \to 2}-\frac{(z-2)^{-1}}{(z-2)^{-2}}$. That looks very good, except the exponents are both negative, meaning that they switch places on the fraction bar.

We now arrive at $\lim\limits_{z \to 2}-\frac{(z-2)^{2}}{(z-2)^{1}}$, and finally $\lim\limits_{z \to 2}(z-2)$. We evaluate the limit directly: $(2-2) = 0$

∴ the lim of $(z-2) \ln|z-2|$ is zero.