Is $f(z)=\exp (-\frac{1}{z^4})$ holomorphic? 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 $\mathbb{C}$ ?
 A: It is neither obvious nor true. Your $f$ has an essential singularity at $z=0$. (what happens if you approach $z=0$ along the line $x=y$ or $x=-y$?)
However, if you split $f = u+iv$, it is true that $u$ and $v$ satisfy Cauchy-Riemann's equations, even at $z=0$. (But $f$ is not continuous at $z=0$, so $u$ and $v$ are certainly not differentiable at $0$.)
See also this answer

Follow-up If $u$ and $v$ are differentiable (as functions $\mathbb{R}^2 \to \mathbb{R}$) and $u$ and $v$ satisfy Cauchy-Riemann's equation at a point $z$ then $f$ is complex-differentiable at $z$. Just assuming $f$ to be continuous is not enough.
However, if we assume that $f$ is continuous on an open set $U$ and that $u$ and $v$ satisfy C-R everywhere on $U$ then $f$ is in fact analytic on $U$, even if we don't assume that $u$ and $v$ are differentiable. This is known as Looman-Menchoff's theorem.
A: 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.
A: Hint: A function is analytic at a point if and only if it has a power series expansion at that point.
