Calculate complex limit converging to constant $e$. I'm stuck for calculating the following complex limit:
$$
\lim_{z \to 0}{(1+z)^{\frac{1}{z}}}
$$
This is the complex version of the real version of limit converging to $e$. (and wolfram says the complex version also converges to $e$) Therefore I guessed that applying the definition of limit ($\epsilon\text{-}\delta$) with $e$ is the way to go.
However I cannot get further. Firstly I guessed that the idea of the proof of the real version of limit might help, but it did not help my brain.
 A: The definition of raising complex numbers to complex powers is given by
$$a^b \equiv e^{b\log a}$$
for some choice of branch cut of $\log$ (when no other information is present, the principal branch of $\log$ is assumed). In this case we have that
$$(1+z)^{\frac{1}{z}} = \exp\left(\frac{\log(1+z)}{z}\right)$$
Since $\exp$ is continuous, we can move the limit in, leading us to find
$$\lim_{z\to 0} \frac{\log(1+z)}{z} = \lim_{z\to 0} \frac{\log(1+z)-\log 1}{z} \equiv (\log z)'\Bigr|_{z=1} = 1$$
which makes the original limit $e$ (as it turns out the branch cut does not matter here because the limit does not need to encircle the origin in $\log$. $\log$ is always differentiable in a neighborhood of $1$).
A: In elementary calculus, the approach to limits having the form $\left[ 1^\infty \right]$ is to use the continuous exponential and logarithm functions to push the base into the exponent then move the limit through the continuous functions.
\begin{align*}
\lim_{z \rightarrow 0} (1+z)^{1/z} &= \lim_{z \rightarrow 0} \exp \ln \left( (1+z)^{1/z} \right)  \\
&= \lim_{z \rightarrow 0} \exp \frac{\ln (1+z)}{z}  \\
&= \exp \lim_{z \rightarrow 0} \frac{\ln (1+z)}{z}
\end{align*}
As you can see this approach has converted the limit to the form $\left[ \frac{0}{0} \right]$, then apply l'Hopital's Rule and finish.
