I'm trying to solve for the limit of the following complex function as $z\to0$. I know L'hopital's rule but I'm to find the answer without using that method. The limit is:
$$\lim_{z \to 0} \dfrac{e^z+z\log (z)}{1-z^2\arg(z)} $$
I know the issue will be with the $\log (z)$ term since $\log(0)$ is not defined. Can anyone see what I should be doing here?
$$L=\lim_{z \to 0} \dfrac{e^z+z\log (z)}{1-z^2\arg(z)} $$ $$L=\dfrac{\displaystyle\lim_{z\to0}e^z+z\log (z)}{\displaystyle\lim_{z\to0}1-z^2\arg(z)}$$ $$L=\dfrac{\displaystyle 1+\lim_{z\to0}z\log (z)}{\displaystyle\lim_{z\to0}1-z^2\arg(z)}$$ For the upper limit, perform change of variables $z\mapsto\frac1z$ $$L=\dfrac{\displaystyle 1+\lim_{z\to\infty}\frac1z\log (\frac1z)}{\displaystyle\lim_{z\to0}1-z^2\arg(z)}$$ $$L=\dfrac{\displaystyle 1-\lim_{z\to\infty}\frac{\log (z)}z}{\displaystyle\lim_{z\to0}1-z^2\arg(z)}$$ Since for any $a>0$ there exists a $Z$ such that for all $z>Z$ it is true that $z>a\log z$. $$L=\dfrac{1}{1-\displaystyle\lim_{z\to0}z^2\arg(z)}$$ $$L=\dfrac{1}{1-\displaystyle\lim_{z\to0}z^2\cdot\lim_{z\to0}\arg(z)}$$ Since $\arg$ as a codomain of $[0,2\pi]$. Note that the notation $[a,b]$ below means that it's some unknown value guaranteed to be inclusively between $a$ and $b$ $$L=\dfrac{1}{1-0\cdot[0,2\pi]}$$ $$L=\dfrac{1}{1-[0,0]}$$ $$L=1$$