Prove a function is not continuous How can I (formally correct) prove that $\;f: \mathbb{C}  \rightarrow \mathbb{C}$ = $ \left\{ \begin{array}{cl} 0 & z = 0 \\ e^{-\frac{1}{z^2}} & z \neq 0 \end{array} \right.$ is not continuous (at $0$)?
 A: A function is continuous at $z_0$ if and only if for every sequence $(z_n)$ that converges to $z_0$, $\lim_{n \rightarrow \infty} f(z_n) = f(z_0)$.
With this, Hint: consider the sequences $z_n = 0$ and $z_n = \sqrt{-1/(2n\pi)}$. (In general, a nice trick is to approach along the real axis, and approach along the imaginary axis.)
A: Hint.  For real $y$, evaluate $\displaystyle\lim_{y\to0} f(iy)$.
A: First of all, define the function as follows:
$
 f(a,b)=
 \begin{cases}
  0                           & \text{$ a =    0 ,\ b =    0$}\\
  e^{-\frac{1}{a^2}}          & \text{$ a \neq 0 ,\ b =    0$}\\
  e^{-\frac{1}{        -b^2}} & \text{$ a =    0 ,\ b \neq 0$}\\
  e^{-\frac{1}{a^2+2abi-b^2}} & \text{$ a \neq 0 ,\ b \neq 0$}\\
 \end{cases}
$
Then, note that $\lim\limits_{b \to 0}{(e^{-\frac{1}{-b^2}})} = \infty \implies \lim\limits_{b \to 0}{f(0,b)} \neq f(0,0) \implies f$ is not continuous at $0,0$.
Moreover, the fact that $\lim\limits_{b \to 0}{f(0,b)} = \infty$ is by itself enough to show that $f$ is not continuous at $0,0$.
