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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.

gnometorule
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Paul
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2 Answers2

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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(0)| = |f(x) - c| >\epsilon $.

Jake Tae
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Ittay Weiss
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  • you're welcome. – Ittay Weiss Mar 04 '13 at 01:29
  • 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 04 '13 at 01:33
  • look at the second part of my answer to see what you need to do. – Ittay Weiss Mar 04 '13 at 01:35
  • 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 04 '13 at 01:40
  • @Paul Firstly, you want to say "with the addendum $f(0) = 0$" not "$1/x^2=0$". The later is never true while the first defines a value for $f$ at $0$. To show that it's discontinuous do what you did above for all δ, so you need to use the variable. Its simple. You know that the function keeps getting bigger near $0$. So, given $\epsilon$, take $x = \frac{1}{2\sqrt{\epsilon}}$ since $f\left(\frac{1}{\sqrt{\epsilon}}\right) = \epsilon$. Now we also need $|x| < \delta$, so take $x = \min\left(\frac{1}{2\sqrt{\epsilon}}, \delta\right)$. Confirm this $x$ works using the inequalities. – Pratyush Sarkar Mar 04 '13 at 03:04
  • I accidentally did it for any $\epsilon$ but you only need to show for one value. For example use $\epsilon = 1$ to make calculations easier. Also when taking $x$, it should say $\delta/2$ in the minimum function above. – Pratyush Sarkar Mar 04 '13 at 03:11
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The stated function $f(x)$ is not defined for $x = 0$ and hence can not be continuous at that point.

Average
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  • So it is then impossible to use the epsilon delta definition to show this? – Paul Mar 04 '13 at 01:25
  • Yes, the $\epsilon -\delta$ definition presupposes that $f(c)$ is defined in the definition of continuity of the function $f(x)$ at $x = c$ – Average Mar 04 '13 at 01:28