Left-hand limit at t = 60

As $t \to 60^-$, we are considering values of $t$ that are slightly less than 60, such as 59, 59.9, 59.99, and so on. For all such values, $t < 60$, so the applicable rule is $g(t) = 5$.

Since $g(t) = 5$ for every value of $t$ in this range, the fee does not change as $t$ gets closer and closer to 60 from below. Therefore $\lim_{t \to 60^-} g(t) = 5$.

As $t \to 60^+$, we consider values of $t$ slightly greater than 60, such as 60.01, 60.001, and so on. For all such values, $t \ge 60$, so the applicable rule is $g(t) = 8$.

Since $g(t) = 8$ throughout this range, the fee stays at 8 as $t$ approaches 60 from above. Therefore $\lim_{t \to 60^+} g(t) = 8$.

From the previous parts, $\lim_{t \to 60^-} g(t) = 5$ and $\lim_{t \to 60^+} g(t) = 8$. These are the left-hand and right-hand limits of $g$ at $t = 60$.

Since a two-sided limit exists only when both one-sided limits are equal, and here $5 \ne 8$, the limit $\lim_{t \to 60} g(t)$ does not exist.

The left-hand limit at t equals sixty was found to be five dollars, while the right-hand limit was found to be eight dollars.

Since eight is greater than five, the right-hand limit is greater than the left-hand limit.

This makes sense because the toll jumps up once the parked time reaches sixty minutes.