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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 continous (at 0)?

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Hi! can you show a bit of your work on this so far? This way it is easier to help you. What's the limit of the exponential for $|z|\to 0$? Does it exist?? –  MattAllegro Apr 28 at 7:40

3 Answers 3

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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 $0,0$.

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I do understand your first point, but why does $\lim_{b\rightarrow 0}f(0,b) = \infty \Rightarrow f$ not continous apply? –  muffel Apr 28 at 10:08
    
@muffel: I thought it was obvious, but now I'm having a bit of a hard time explaining it to myself... Anyway, suppose $\lim\limits_{b \to 0}{f(0,b)} = f(0,0)$, then $f(0,0) = \infty$, hence $f$ is not even defined at $0,0$ (let alone continuous at that point). –  barak manos Apr 28 at 10:41
    
Basically it's because the definition of "continuous at $a$" is the same as $\lim\limits_{z \rightarrow a}f(z) = f(a)$, once you get down to deltas and epsilons. If $f$ diverges to infinity along one "line of approach" to $0$, then that limit doesn't exist let alone is equal to whatever value you've chosen for $f(0)$. –  Steve Jessop Apr 28 at 12:25
    
@Steve Jessop: So isn't that essentially what I wrote in the comment above? –  barak manos Apr 28 at 12:32
    
@barak: sure, I'm just expressing it in slightly different terms since you said you found it difficult to explain. I think it might help muffel. –  Steve Jessop Apr 28 at 12:33

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.)

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

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