Epsilon delta proofs on discontinuous functions

Question

Does the epsilon delta definition of a limit always hold? I was thinking about the function $$f(x)=\begin{cases} 1 & x= 0 \\ 0 & x\neq 0 \end{cases} $$ and it seems that $$\lim_{x \to 0} f(x) = 0$$, however unless I'm mistaken this would not converge to $0$ (or even at all) via the epsilon delta definition as given $$\epsilon \geq 1, \nexists \space \delta>0 \space s.t. \space|x|<\delta \implies |f(x)|<\epsilon$$ as we have $f(0)=1$. There's probably just something I'm missing but it's bothering me.

Answer 1

If $\lim\limits_{x \rightarrow 0}f(x) = 0$ , then given $\epsilon>0$, there must be $\delta>0$ such that $0<|x-0|<\delta, x \neq 0 \implies |f(x)-0|<\epsilon$

Now, for $x \neq 0$, $f(x)=0$, then for any $\delta>0$

$0<|x-0|<\delta \implies |0-0|<\epsilon \implies |f(x)-0|<\epsilon$.