1. **State the problem:** We want to find the long-term amount of warfarin in the body, given that each day a 5 mg dose is taken and 8% of the drug remains after 24 hours. 2. **Define the sequence:** The amount of warfarin before the $(n+1)$st dose is given by $$Q(n) = 5 \times \frac{8}{100} \times \left(1 + \frac{8}{100} + \left(\frac{8}{100}\right)^2 + \cdots + \left(\frac{8}{100}\right)^{n-1}\right)$$ 3. **Recognize the sum:** The expression inside the parentheses is a geometric series with first term $a = 1$ and common ratio $r = \frac{8}{100} = 0.08$. 4. **Formula for geometric series sum:** For $|r| < 1$, the sum of the first $n$ terms is $$S_n = \frac{1-r^n}{1-r}$$ 5. **Apply the formula:** Substitute $r=0.08$ into the sum: $$Q(n) = 5 \times 0.08 \times \frac{1 - (0.08)^n}{1 - 0.08}$$ 6. **Simplify denominator:** $$1 - 0.08 = 0.92$$ 7. **Simplify the expression:** $$Q(n) = 5 \times 0.08 \times \frac{1 - (0.08)^n}{0.92} = \frac{0.4}{0.92} \times \left(1 - (0.08)^n\right)$$ 8. **Calculate the limit as $n \to \infty$:** Since $0 < 0.08 < 1$, we have $$\lim_{n \to \infty} (0.08)^n = 0$$ Therefore, $$\lim_{n \to \infty} Q(n) = \frac{0.4}{0.92} \times (1 - 0) = \frac{0.4}{0.92}$$ 9. **Final calculation:** $$\frac{0.4}{0.92} \approx 0.4348$$ **Answer:** The long-term amount of warfarin in the body is approximately **0.4348 mg** before the next dose.