Isolated singularity $0$

Question

Let $f:D^*_1(0) \to \mathbb{C}$ be analytic function such that:

$$\lim_{n \to \infty} 3^nf(\frac{1}{2^n})=1$$

What is the nature of the isolated singularity $0$?

I know that $$\lim_{n \to \infty} 3^nf(\frac{1}{2^n})=1 \implies \lim_{n \to \infty} f(\frac{1}{2^n})= \lim_{n \to \infty} \frac{1}{3^n}=0$$ it suggests to me that $ 0 $ is removable.So I want to prove that $$\lim_{z \to 0}f(z)=0$$ I can't argue to prove that this limit is actually zero.

Answer 1

Since $\lim_{n \to \infty} 3^nf(\frac{1}{2^n})=1$, there is $N$ such that

$$\frac{1}{2 \cdot 3^n} \le f(\frac{1}{2^n}) \le \frac{2}{ 3^n}$$

for $n >N.$ This gives $f(\frac{1}{2^n}) \to 0$ as $n \to \infty.$ Therefore $0$ is not a pole of $f$.

Now assume that $0$ is a removable singularity of $f$. Since $f(\frac{1}{2^n}) \to 0$ as $n \to \infty,$ we have $f(0)=0.$ Hence there is $m \in \mathbb N$ and a holomorphic $g$ with

$$f(z)=z^mg(z)$$

and $g(0) \ne 0$. Then we derive

$$3^nf(\frac{1}{2^n})= (\frac{3}{2})^n \frac{1}{2^m}g(\frac{1}{2^n}).$$

Since $\lim_{n \to \infty} 3^nf(\frac{1}{2^n})=1$, we see that $g(\frac{1}{2^n})) \to0$, thus $g(0)=0,$ a contradiction.

Consequence: $f$ has in $0$ an essential singularity.