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I have the function $f(x) = \frac{1}{x^2}$. I want to show that it is discontinuous at $x=0$ using the epsilon delta definition.

So, I need to show that for all $\epsilon > 0$, there does not exist a $\delta >0$, s.t. $|x|<\delta$ $\Rightarrow |\frac{1}{x^2} - \frac{1}{0^2}|$.

However, I do not see how to proceed with this inequality when $\frac{1}{0^2}$ is simply undefined. I think this is straightforward to do with proving functions are continuous at points, but I do not see how to do it with discontinuities.

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Are you sure that's what you need to prove? The function you have is not defined at zero to begin with... – Sara Mar 4 '13 at 1:25

A function that is not defined at a point $x$ is automatically not continuous at that point. So, there is actually nothing to show in order to prove what you want, except noting that the point $x=0$ is not in the domain of definition of the function.

More interestingly, if you define $f(0)=c$, then no matter what $c$ is the function is discontinuous at $x=0$. You can show that by showing that there exists an $\epsilon > 0$ such that for all $\delta >0$ there exists $x$ such that $|x|<\delta$ but $|f(x)-f(c)|>\epsilon $.

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Thank you for the reply. – Paul Mar 4 '13 at 1:28
you're welcome. – Ittay Weiss Mar 4 '13 at 1:29
I am now trying to work through the same problem with the addendum that $1/x^2$ = 0. Then, I need to show that |x| < delta => $|1/x^2|$ > epsilon. I am not sure how to proceed. How do I find the specific epsilon that exists? – Paul Mar 4 '13 at 1:33
look at the second part of my answer to see what you need to do. – Ittay Weiss Mar 4 '13 at 1:35
That is how I came to $|x| < delta$ => $|1/x^2| > epsilon$, but I see no connection between the two except arbitarily picking values of epsilon and delta. I could say there exists x s.t. $|x| < 0.25$ but $|1/x^2|$ > .1. – Paul Mar 4 '13 at 1:40

The stated function $f(x)$ is not defined for $x = 0$ and hence can not be continuous at that point.

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So it is then impossible to use the epsilon delta definition to show this? – Paul Mar 4 '13 at 1:25
Yes, the $\epsilon -\delta$ definition presupposes that $f(c)$ is defined in the definition of continuity of the function $f(x)$ at $x = c$ – newToProgramming Mar 4 '13 at 1:28

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