Consider the sequence of functions $$f_n(z) = \begin{cases} n, & \text{if $0<|z|\leq\frac{1}{n}$} \\ \frac{1}{z^4}, & \text{if $\frac{1}{n}<|z|\leq1$} \end{cases} $$ for $n\geq 1$, on the closed, punctured disk, $\overline{D}\prime(0,1) = \{z \in \mathbb{C} : 0<|z|\leq1\}$.
(i) Prove that $f_n(z)$ converges pointwise to a functions $f(z)$ and determine this function.
(ii) Is the convergence uniform?
For part (i) I've guessed that $f(z) = \frac{1}{z^4}$ and I am trying to prove this by showing that $\lim_{n \to \infty}|f_n(z)-f(z)| = 0$. I've managed to do this for the case when $\frac{1}{n}<|z|\leq1$ as $\lim_{n \to \infty}|f_n(z)-f(z)| = \lim_{n \to \infty}|\frac{1}{z^4}-\frac{1}{z^4}| = 0$ in that case. However, for the case when $0<|z|\leq\frac{1}{n}$ I haven't had much luck. Thus far for this case I have that $$\lim_{n \to \infty}|f_n(z)-f(z)| = \lim_{n \to \infty}\left|n-\frac{1}{z^4}\right| \leq n + \frac{1}{|z|^4}$$ and as $|z| \leq \frac{1}{n}$ in this case we have that $n-\frac{1}{|z|^4} \leq n-n^4$. But I've no ideas about where to from here since $n-n^4 \to -\infty$ as $n \to \infty$.
I've also made an attempt at part (ii) but haven't had any luck with that either.
Thank you in advance.
It is sufficient to note that for $n\to \infty$, the interval $0 < |z| \le 1/n$ "crumbles" to the point $0$, where the function is undefined.
In particular, given $z\in \Bbb{C}$, you have that: $$n > 1/|z| \implies |f_n(z) - z^{-4}| = 0$$ So the sequence converges pointwise. So for any $z$ we can find an $n$ such that for all subsequent $n$, the two functions are the same.
EDIT: It does not converge uniformly, since for all $M>0$: $$n \ge 1 \implies \sup_{z \in D} |f_n(z) - z^{-4}| \ge M$$ As you can clearly see, since the supremum of $f_n$ over the disk is $n^4$ while the supremum $z^{-4}$ is infinite. The absolute value of their difference is, again, infinite in supremum.