Answer
$\lim\limits_{t \to 0} H(t)$ does not exist.
Work Step by Step
$H(t) = 0$, if $t \lt 0$
$H(t) = 1$, if $t \geq 0$
Assume $\lim\limits_{t \to 0} H(t) = c$
Let $\epsilon = \frac{1}{2}$
Choose any value for $\delta \gt 0$
Choose a number $a$ such that $-\delta \lt a \lt 0$
Choose a number $b$ such that $0 \lt b \lt \delta$
$\vert a - 0 \vert \lt \delta$ and $\vert b - 0 \vert \lt \delta$
Then $\vert H(a) - c \vert \lt \epsilon = \frac{1}{2}$ and $\vert H(b) - c \vert \lt \epsilon = \frac{1}{2}$
Since $H(a) = 0$, then $-\frac{1}{2} \lt c \lt \frac{1}{2}$
Since $H(b) = 1$, then $\frac{1}{2} \lt c \lt \frac{3}{2}$
Clearly this is a contradiction since $c$ can not be in both of these intervals.
Therefore, our assumption that $\lim\limits_{t \to 0} H(t) = c$ is false. Thus $\lim\limits_{t \to 0} H(t)$ does not exist.
