TITLE: Is $f(z)=\exp (-\frac{1}{z^4})$ holomorphic?
QUESTION [6 upvotes]: Let $f(z)=\exp (-\frac{1}{z^4})$ for $z\neq 0$ and $f(0)=0$. 

Is it obvious that 
  $$\lim_{z\to z_{0}}\frac{f(z)-f(0)}{z-0}=\lim_{z\to z_{0}}\frac{\exp (-\frac{1}{z^4})}{z}=0$$ 

And if this limit is indeed equal to $0$; can we conclude that $f$ is holomorphic on $\mathbb{C}$ ?

REPLY [2 votes]: If you define a sequence $ (z_{n})_{n \in \mathbb{N}} \stackrel{\text{def}}{=} \left( \dfrac{\sqrt[4]{i}}{n} \right)_{n \in \mathbb{N}} $, then $ |f(z_{n})| = \left| e^{i n^{4}} \right| = 1 $ for all $ n \in \mathbb{N} $. Hence, we do not have $ \displaystyle \lim_{n \to \infty} f(z_{n}) = 0 $, even though $ \displaystyle \lim_{n \to \infty} z_{n} = 0 $. Here, $ \sqrt[4]{i} $ denotes an arbitrary fourth root of $ i $.
Conclusion: $ f $ is not continuous at $ z = 0 $, much less complex-differentiable there.