Find out the limit of $f(z)=(z-2)\log|z-2|, z\neq 2$ at the point $z_0=2$, or explain why it does not exist. 
Question: Find out the limit of $f(z)=(z-2)\log|z-2|, z\neq 2$ at the point $z_0=2$, or explain why it does not exist.

My approach: Let $z=x+iy$, where $x,y\in\mathbb{R}$. This implies that $f(z)=f(x+iy)=(x+iy-2)\log|x+iy-2|=(x+iy-2)\log\left(\sqrt{(x-2)^2+y^2}\right).$
Therefore, $u:=\Re(f(z))=(x-2)\log\left(\sqrt{(x-2)^2+y^2}\right)$ and $v:=\Im(f(z))=y\log\left(\sqrt{(x-2)^2+y^2}\right).$
Now $$\lim_{z\to 2}u=\lim_{(x,y)\to (2,0)}u(x,y)=\lim_{(x,y)\to (2,0)}(x-2)\log\left(\sqrt{(x-2)^2+y^2}\right)$$ and $$\lim_{z\to 2}v=\lim_{(x,y)\to (2,0)}v(x,y)=\lim_{(x,y)\to (2,0)}y\log\left(\sqrt{(x-2)^2+y^2}\right).$$ Observe that both of these limits are of the indeterminate form $0.(-\infty)$. But, I cannot use L'Hopital's rule here, and I do not know if there is a multi-variable form of L'Hopital's rule or not.
So, how to proceed after this and is there any other way to solve the problem?
 A: By definition of limit the statement  $$(z-2)\ln |z-2| \to 0$$ as $z \to 2$ is equivalent to the statement $$t \ln t \to 0$$ as $t \to 0+$. Do you know how to prove this?
[We have to show that given $\epsilon >0$ there exists $\delta >0$ such that $|(z-2)\ln |z-2|| <\epsilon$ whenever  $0<|z-2 | <\delta$. Just Put $t =|z-2|$ to see the equivalence. The fact that $t \ln t \to 0$ is proved by applying L'Hopital's Rule to $\frac {\ln t} {1/t}$].
A: *

*We have $|f(z)|= \log( |z-2|^{|z-2|}).$


*$x^x \to 1$ as $x \to 0^+.$
Consequence: $|f(z)| \to \log 1=0$ as $z \to 2.$ Therefore
$$f(z) \to \log 1=0$$
as $z \to 2.$
A: Claim: $\lim_{z\to 2}f(z)=\lim_{z\to 2}(z-2)\log|z-2|=0$.
Proof: By the definition of limit of a complex-valued function of a complex variable, if we can show that, given any $\epsilon>0$, there exists a $\delta >0$, such that $\forall z$  satisfying $0<|z-2|<\delta$ and $z$ belonging to the domain of definition of the function $f$ (in this case $z\neq 2$), we have $$|f(z)-0|<\epsilon, \text{ i.e}, |(z-2)\log|z-2||<\epsilon,$$ then we can conclude that $$\lim_{z\to 2}(z-2)\log|z-2|=0.$$
Now setting $|z-2|=t$, we have $|(z-2)\log|z-2||=|z-2||\log|z-2||=t|\log t|.$
Thus, the above mentioned $\epsilon-\delta$ statement is equivalent to the statement which states that, given any $\epsilon>0$, there exists a $\delta>0$, such that $\forall t$ satisfying $0<t<\delta$, we have $$t|\log t|<\epsilon.$$
Therefore, if we can show that $$\lim_{t\to 0^+}t|\log t|=0,$$ then we will be done.
Now $$t|\log t|=\begin{cases} t\log t, \hspace{0.5 cm}t\ge 1 \\ -t\log t, \hspace{0.1 cm}0<t<1.\end{cases}$$
Thus, $\lim_{t\to 0^+}t|\log t|=\lim_{t\to 0^+}-t\log t=\lim_{t\to 0^+}\frac{\log t}{-1/t}=\lim_{t\to 0^+}\frac{1/t}{1/t^2}=\lim_{t\to 0^+}t=0.$
Hence, we are done and we have $$\lim_{z\to 2}(z-2)\log|z-2|=0.$$
