# Show using the epsilon delta definition that this function is discontinuous at 1

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}

• Take $\epsilon=1/2$ and see what happens. – Michael Hoppe Nov 5 '18 at 18:59
• What does "x =/=0" mean? – John Douma Nov 5 '18 at 19:01
• Possible duplicate of proving continuity using epsilon delta – Don Thousand Nov 5 '18 at 19:02
• It means $x \neq 0$. – KM101 Nov 5 '18 at 19:02

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 $$\bigl\lvert f(x)-1\bigr\rvert\geqslant\varepsilon$$.
• You can't impose any condition on $\delta$. Did you not read what I wrote? It says “for every $\delta>0$”. – José Carlos Santos Nov 5 '18 at 19:24
• I don't see what you mean by “general”, but, yes, you have to find a $x$ for each $\delta>0$. – José Carlos Santos Nov 5 '18 at 19:55