Is $f(z)$ entire?

Question

I am trying to determine if the the following is entire $$f(z)= \begin{cases} e^{-z^{-4}} & z\neq0 \\ 0 & z=0\\ \end{cases} $$

My attempt:

Consider $z\ne 0$. $f(z)=e^{-z^{-4}}$ is the composition of two differentiable functions, hence $f$ is differentiable when $z\neq 0$.

Next I considered $z=0$. The Cauchy-Riemann equations are satisfied at $z=0$, as $u_x=u_y=v_x=v_y=0$. But this is not sufficient condition for complex differentiability. I wish to show $f$ is continuous/discontinuous at $z=0$.

I tried defining the limit $$f(z)=\lim_{z\to 0} e^{-z^{-4}}=e^{-\infty}=0$$

This would imply $f$ is continuous at $z=0$, but the answer I have disagrees.

edit

If I let $z=re^{i\theta}$, then \begin{align} \lim_{z\to 0} e^{-z^{-4}}&=\lim_{r\to 0} e^{-\frac{1}{r^4}e^{-4i\theta}} \\ &=\lim_{r\to 0} e^{\frac{1}{r^4}}\rightarrow \infty \ \ \ \ \text{letting} \left(\theta=\frac{\pi}{4}\right) \\ \end{align} Hence, $f$ is not continuous at $z=0\implies f$ is not differentible at $z=0$.

So $f$ is not entire.

Answer 1

Hint: consider $z=x+xi$ for real $x\to0$.

Answer 2

For each positive interger $n$ lat $c_n$ be a fourth root of $ {2ni\pi} $ and $z_n = \frac 1 {c_n}$Then $|e^{-z_n^{-4}} |=1$ for all $n$ and $z_n \to 0$.