For $u \in \mathbb C \setminus (-\infty, 0]$ and $p \in \mathbb R$ define $$\operatorname{Log} u = \int_{<1, u>}\frac {dz}{z}, \quad u^p = \exp(p \operatorname{Log}(u)).$$
On this domain, $f(u) = u^p$ is holomorphic. Does the limit at $0$ exist?
At first I thought that for positive $p$ it must be zero, but then after looking at the definition I noticed that there is exponential of something approaching complex infinity, which is not good.
Context: I was trying to calculate characteristic function of gamma distribution, and after change of variables I ended up with complex integral. I wanted to use the Cauchy theorem, to say that it is equal to gamma function (the same function integrated over real positive line instead of some complex line), but I'm not sure if the part near zero converges; for that I would need $\lim_{u \to 0} u^{p-1} = 0$, for any positive $p$.
Converting my comment to an answer: you have $\log(u)=\log(|u|)+i\arg(u)$, and as $u\to 0$, the real part approaches $-\infty$. This minus sign, together with the assumption $p>0$, implies that the limit in question is $0$, because the imaginary part plays no role ($e^{ix}$ is bounded for real $x$).
However, $\lim\limits_{u\to 0}u^p=\infty$ (in the complex sense) if $p<0$, and this follows because now the real part of $p\log(u)$ approaches $+\infty$. So, your last sentence isn’t quite true.
For completeness sake: when $p=0$, the function you get is identically $1$ on $\Bbb{C}\setminus(-\infty,0]$, so it has a continuous extension to all of $\Bbb{C}$, which is still the constant function $1$.