Prove a function is not continuous

Question

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 $0$)?

Answer 1

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 $0,0$.

Answer 2

Hint. For real $y$, evaluate $\displaystyle\lim_{y\to0} f(iy)$.

Answer 3

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.)