Evaluate $\lim_{z\to\infty} \frac{(1 - z^4)^{1/4}}{z}$.

Question

I'm going through my (applied) complex analysis notes from almost a year ago and trying to make sense of something I wrote down. Of course, anybody's own solution is very much welcome.

We used the "+"-shaped branch cut, $\{z ~|~ z \in [-1,1] \subseteq \mathbb{R}\} \cup \{z ~|~ \text{Re}(z)=0,\, \text{Im}(z) \in [-1,1]\}$, with the specified branch of $(1-z^4)^{1/4}$ corresponding to the first quadrant of the plane evaluating to the positive fourth root. Here is what I have written down:

The possible values are $e^{i\pi/4}, -e^{i\pi/4}, ie^{i\pi/4}, -ie^{i\pi/4}$. Again, relying on continuity of the branch, what is the best path to $\infty$?

Choose $1 + \epsilon \to \infty$ along the positive real axis. Factoring the numerator,

\begin{align*} (1 - z^4) &= (-1)(z - 1)(z + 1)(z - i)(z + i) \\ (1 - z^4)^{1/4} &= (-1)^{1/4}(z - 1)^{1/4}(z + 1)^{1/4}(z - i)^{1/4}(z + i)^{1/4}. \end{align*}

Let $z = 1 + w$ with $w$ small, i.e. $w = \epsilon e^{i\theta}$, then

$$ f(1 + w) = (-1)^{1/4} (w^{1/4})(2 + w)^{1/4} (2 + 2w + w^2)^{1/4} $$

for $\theta \in (-\pi,\pi)$. For $\theta = \pi$, then

$$f(z) \approx \sqrt{2} w^{1/4} e^{-i\pi/4} + (\text{smaller terms}).$$

The argument of $f(z)$ for $z = 1 + w$ for small $w$ is $-\pi i/4$. The limit is therefore $e^{-i\pi/4} = -ie^{i\pi/4}$.

I very much remember not comprehending it at the time as well many other classmates, and also remember that the professor stated that they made an error near the end because he forgot the $\sqrt{2}$, so I may have added that back in wrong. I apologize for any lack of clarity. My questions are

1. If we're taking a path along the positive real axis as in $1 + \epsilon \to \infty$, then why are we taking $\epsilon \to 0$ as in "$w$ small" in the computation?

2. Why choose $\theta = \pi$. My guess for the latter is so that the your values "remain" in the first quadrant because your dividing the argument of $e^{\pi i}$ by 4.

3. And the major question is that I can't figure out why with $|w|$ small that $\arg f(z) = -\pi i/4$. Why negative? I'm guessing it's due to how we evaluate $(-1)^{1/4} = (e^{-\pi i})^{1/4} = (e^{\pi i})^{1/4}$ but that confuses me if we're also taking $w = \epsilon e^{i\theta}$ and evaluating $\theta = \pi$.

4. (Edit:) Another question: Is the statement of the problem, well, problematic? Does this problem even make sense? Does clarifying the limit to mean as $|z| \to +\infty$ help at all?