The Shadow Revenue Limit
A campus coffee cart models daily revenue from a new drink as R(t) = 120·(1 − e^(−0.4t))/t dollars, where t is the number of weeks since launch (t > 0). At launch week t = 0 the formula is undefined, producing a 0/0 form, but the owner needs a sensible "launch-day revenue rate" to justify the investment. A partner argues that because the formula blows up at t = 0, revenue is infinite on day one and the model is useless. The owner instead suspects the limit is finite and wants it computed two independent ways to be sure. She also wants to know whether the revenue is still climbing or leveling off after many weeks.
1. Evaluate lim(t→0⁺) R(t) without a calculator shortcut, showing that the 0/0 form resolves to a finite value. Use a series or standard limit argument.
Read the worked answer
Write e^(−0.4t) = 1 − 0.4t + (0.4t)²/2 − … Then 1 − e^(−0.4t) = 0.4t − 0.08t² + …
Divide by t:
R(t) = 120·[0.4 − 0.08t + …] = 48 − 9.6t + …
As t → 0⁺, R(t) → 48 dollars per week. The 0/0 form is removable: the numerator vanishes linearly, exactly canceling the t in the denominator. The partner is wrong - the singularity is removable, not infinite. (Equivalently, by L'Hôpital: lim 120·(0.4e^(−0.4t))/1 = 120·0.4 = 48.)
2. Find lim(t→∞) R(t) and interpret it. Does the model predict revenue keeps growing without bound, levels off, or decays to zero? Justify with the behavior of the numerator.
Read the worked answer
As t → ∞, e^(−0.4t) → 0, so the numerator 120(1 − e^(−0.4t)) → 120, a finite nonzero constant. Dividing a constant near 120 by a growing t gives
lim(t→∞) R(t) = 0.
So the per-week revenue rate decays to zero: the drink's weekly revenue peaks near launch (about $48/week) and then tapers off as the one-time novelty fades, even though cumulative revenue keeps accumulating. Note the contrast with the t→0 limit: the function is finite at both ends (48 at 0⁺, 0 at ∞) with no vertical asymptote - the only genuine feature is the removable hole at t = 0.