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The function to be studied is the following one: $$ f(x)= \begin{cases} x^2 &\text{ when } x \neq 0\\ 1 &\text{ when } x = 0 \end{cases} $$ Show that $f(x)$ is discontinuous at x=0.
How do I prove this strictly from the epsilon delta definition? I believe that this means that there exists an epsilon such that $$ |f(x)-1|< \epsilon, $$ where there is no such delta for which $$ |x-0|< \delta. $$But I'm unsure how to continue using the definition to prove this.

I have solved this question, taking x = Min{1/2, δ}

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  • $\begingroup$ Take $\epsilon=1/2$ and see what happens. $\endgroup$ – Michael Hoppe Nov 5 '18 at 18:59
  • $\begingroup$ What does "x =/=0" mean? $\endgroup$ – John Douma Nov 5 '18 at 19:01
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    $\begingroup$ Possible duplicate of proving continuity using epsilon delta $\endgroup$ – Don Thousand Nov 5 '18 at 19:02
  • $\begingroup$ It means $x \neq 0$. $\endgroup$ – KM101 Nov 5 '18 at 19:02
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Asserting that $f$ is continuous at $0$ means that, for every $\varepsilon>0$, there is some $\delta>0$ such that $\lvert x\rvert<\delta\implies\bigl\lvert f(x)-f(0)\bigr\rvert<\varepsilon$. Therefore, asserting that $f$ is discontinuous at $0$ means that there is some $\varepsilon>0$ such that, for every $\delta>0$, there is a number $x$ such that $\lvert x\rvert<\delta$ and that $\bigl\lvert f(x)-f(0)\bigr\rvert\geqslant\varepsilon$. Take $\varepsilon=\frac12$ and prove that, indeed, for every $\delta>0$, there is a number $x$ such that $\lvert x\rvert<\delta$ and $\bigl\lvert f(x)-1\bigr\rvert\geqslant\varepsilon$.

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  • $\begingroup$ is x = 𝛿/2 likely to work, its clear that then |x| < 𝛿, could I then impose a condition on 𝛿, so that I would have x^2 < 1/2? Would that then leave me with |x^2 - 1| > 1/2 if that condition was the case? $\endgroup$ – Suiri Nov 5 '18 at 19:19
  • $\begingroup$ You can't impose any condition on $\delta$. Did you not read what I wrote? It says “for every $\delta>0$”. $\endgroup$ – José Carlos Santos Nov 5 '18 at 19:24
  • $\begingroup$ oh my mistake, the only condition I'm allowed to have is that 𝛿 > 0 $\endgroup$ – Suiri Nov 5 '18 at 19:36
  • $\begingroup$ I have to find a general x for every 𝛿? $\endgroup$ – Suiri Nov 5 '18 at 19:45
  • $\begingroup$ I don't see what you mean by “general”, but, yes, you have to find a $x$ for each $\delta>0$. $\endgroup$ – José Carlos Santos Nov 5 '18 at 19:55

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