I made a mistake with $e$ in $\mathbb{C}$ , but I ammended it

( $e^z \sim 1 + z$ , as $z \to 0$ ), and ( $\ln (1 + z) \sim z$ , as $z \to 0$ ), so $\displaystyle \lim_{z \to 0} (e^{z})^{\frac{1}{z}} = \lim_{z \to 0} (e^{\frac{1}{z} (\ln (1 + z))}) = \lim_{z \to 0} e^{(\frac{1}{z} \cdot z)} = e$ What about the bonus ? Warning: You can't use my previous argument .

Bonus . $\displaystyle \lim_{z \to 0} (e^{\frac{1}{z}})^{z} = \lim_{z \to \infty} (e^z)^{\frac{1}{z}}$ This limit doesn't exist in $\mathbb{C}$ , because if $R \in \mathbb{R}, R >0$ , $\displaystyle \lim_{R \to 0^{+}} (e^{\frac{1}{R}})^{R} = \lim_{\frac{1}{R} \to + \infty} (e^R)^{\frac{1}{R}} = e$ , and taking $iR, \text{ such that } R \in \mathbb{R}, R >0$ , $\displaystyle \lim_{R \to 0^{+}} (e^{\frac{1}{iR}})^{iR} = \lim_{R \to 0^{+}} e^{iR(ik)} = 1$ due to $k \in (\pi, \pi]$ and $k$ is the main argument of $e^{\frac{1}{iR}}$ , ( $\text{ Arg }e^{\frac{1}{iR}}$ ), $\ln e^{\frac{1}{iR}} = \ln e^{\frac{-i}{R}}= \ln 1 + ik$