Let $H(t)$ be the heaviside function
$$H(t) = \begin{cases} 1, \quad &t\ge 0\\ 0, \quad &t<0\end{cases}.$$
Prove that $\lim_{t\to 0}H(t)$ does not exist using $\varepsilon$ - $\delta$ definition.
Hint: let $\varepsilon = \frac 12$ and suppose $\lim_{t\to 0}H(t) = L$
I have in my mind that I should separate this into cases, and those are when $t \lt 0$ or $t \ge 0$, but I can't formalize it the way the proof requires. My attempt was the following:
Suppose $\lim_{t\to 0}H(t) = L$
$\forall \varepsilon \gt 0$, $\exists \delta \gt 0$ s.t.
$0 \lt |t| \lt \delta$ $\implies$ $|H(t) - L| \lt \frac 12$
If $t$ aproaches $0$ from the right, we would have $|1 - L| \lt \frac 12$, and $|0 - L| \lt \frac 12$ if it's coming from the left. From here I have no idea.
Suppose that $\lim_{t \to 0} H(t)=L$. Then by definition, for any $\varepsilon>0$, we can find a $\delta>0$ such that $$|t-0|<\delta \implies |H(t)-L|<\varepsilon$$ If this is true for any epsilon, let's try $\varepsilon=\frac{1}{2}$.
Case 1: Suppose $t>0$. Then $H(t)=1$. So by assumption, $\exists\delta>0$ such that $$|t|<\delta \implies |1-L|<\frac{1}{2}$$
Case 2: Suppose $t<0$. Then $H(t)=0$. So by assumption, $\exists\delta>0$ such that $$|t|<\delta \implies |0-L|<\frac{1}{2}$$
There is no number $L$ such that $|1-L|<\frac{1}{2}$ and $|0-L|<\frac{1}{2}$ are both true. Can you explain why? Therefore, we have reached a contradiction, and our original assumption is false.