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}
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 $\bigl\lvert f(x)-1\bigr\rvert\geqslant\varepsilon$.