Application of Limit Function

Application in Public Health Problems

One of the most relevant applications of limit functions is in disease spread analysis and vaccination programs. When governments design public health strategies, they need to understand how the number of cases will change over time and how many resources are needed.

Virus Spread Model

Suppose in a city there is a function that describes the number of residents infected with a virus:

where $N(t)$ represents the number of infected residents and $t$ represents time in certain units.

To understand the long-term behavior of this spread, we need to calculate:

Long-term behavior analysis:

When $t$ is very large, the term $t^2$ will dominate inside the root because:

• $t^2$ grows faster than $t$ and the constant $(190.68)^3$
• For $t >> 1$, then $t^2 - t + (190.68)^3 \approx t^2$

So,

This negative result mathematically indicates that this model is only valid for a limited time period. In real context, the number of infected residents cannot be negative, so this model is only valid up to the point where $N(t) \geq 0$.

Optimal Vaccination Strategy

In the context of vaccination programs, limit functions help determine effective vaccination targets. If we know that at a certain time the number of cases will stabilize or decrease, we can calculate how many vaccines are needed.

Suppose the vaccination target for residents over 18 years old is $V$ people, and we want to achieve a condition where the number of new cases approaches zero. We can use limits to determine the optimal strategy.

Application in Economics

Marginal Cost Analysis

In economics, marginal cost is the additional cost to produce one additional unit. Practically, this answers the question: "How much additional cost if we produce 1 more unit?"

Mathematical definition using limits:

where:

• $C(x)$ = total production cost function for $x$ units
• $\Delta x$ = small change in production quantity
• The limit gives the instantaneous rate of change of cost with respect to production

Population Growth Model

In demographic studies, population growth models often use functions involving limits. For example, the logistic model:

The limit of this function when $t \to \infty$ gives the environmental carrying capacity:

Application in Technology and Science

Digital Signal Analysis

In digital signal processing, limit functions are used to analyze system response to certain inputs. Digital filters are often evaluated using limits to understand high and low frequency behavior.

Chemical Reaction Rate

In chemistry, reaction rates can be modeled using exponential functions. Limit functions help determine the equilibrium concentration of reactants:

City Vaccination Program Optimization

1. Problem Description:

   A city with a population of 576,260 people is facing a disease outbreak. The city government has developed a mathematical model to predict the number of positive cases based on the number of people who have been vaccinated. The model is expressed in the function:

   $$N(t) = 285000 - \sqrt{t^2 - t + (190.68)^3}$$

   where:

   • $N(t)$ = number of positive cases remaining
   • $t$ = number of people who have been vaccinated

2. Question:

   If the vaccination program target is 282,367 people, how many positive cases will remain when the target is reached?

3. Data and Assumptions:

   Based on the city's demographic survey:

   • Total population: 576,260 people
   • Age composition: 21% children (≤18 years), 79% adults (>18 years)
   • Health status: 30% already confirmed positive
   • Policy: Only adults who are not positive can be vaccinated

   Justification for Vaccination Target of 282,367 people:

   $$\text{Children} = 0.21 \times 576{,}260 = 121{,}015 \text{ people}$$

   $$\text{Adults} = 0.79 \times 576{,}260 = 455{,}245 \text{ people}$$

   $$\text{Total positive} = 0.3 \times 576{,}260 = 172{,}878 \text{ people}$$

   $$\text{Positive adults} = 0.79 \times 172{,}878 = 136{,}574 \text{ people}$$

   $$\text{Adults eligible for vaccination} = 455{,}245 - 136{,}574 = 318{,}671 \text{ people}$$

   The vaccination target of 282,367 people represents 89% of eligible adults, adjusted for vaccine availability and medical staff capacity.

4. Solution:

   To determine the number of cases remaining when $t = 282{,}367$, we calculate:

   $$N(282{,}367) = 285{,}000 - \sqrt{(282{,}367)^2 - 282{,}367 + (190.68)^3}$$

   Step 1: Calculate $(190.68)^3$

   $$(190.68)^3 = 190.68 \times 190.68 \times 190.68 = 6{,}932{,}907.88$$

   Step 2: Calculate $(282{,}367)^2$

   $$(282{,}367)^2 = 79{,}731{,}122{,}689$$

   Step 3: Substitute into the root

   $$79{,}731{,}122{,}689 - 282{,}367 + 6{,}932{,}907.88 = 79{,}737{,}773{,}229.88$$

   Step 4: Calculate the final result

   $$N(282{,}367) = 285{,}000 - \sqrt{79{,}737{,}773{,}229.88}$$

   $$= 285{,}000 - 282{,}378.78$$

   $$= 2{,}621.22 \approx 2{,}621 \text{ people}$$

5. Result Interpretation:

   When the vaccination program reaches the target of 282,367 vaccinated people, the model predicts that there will be 2,621 positive cases remaining that still need to be handled. This result provides important information for subsequent health resource planning.

Interpretation and Decision Making

Understanding limit results is very important in decision making. In the example above, the result of 2,621 cases provides information to policymakers about:

1. Hospital capacity still needed
2. Number of medical personnel that must be prepared
3. Resource allocation for handling remaining cases
4. Communication strategy to the public about realistic expectations

Limit functions provide insights into the long-term behavior of systems, enabling more effective and realistic planning.

Exercises

1. A pharmaceutical company models vaccine production with the function $P(t) = 50000(1 - e^{-0.1t})$. Determine the maximum production capacity using the concept of limits.

2. The function of information spread on social media is expressed as $I(t) = \frac{100000t}{t + 50}$. Calculate the limit when $t \to \infty$ and interpret the result.

3. The total cost of mask production is $C(x) = 1000 + 5x + 0.01x^2$. Determine the marginal cost using the limit definition.

Answer Key

1. Solution:

   Maximum production capacity is obtained by calculating the limit when $t \to \infty$:

   $$\lim_{t \to \infty} P(t) = \lim_{t \to \infty} 50000(1 - e^{-0.1t})$$

   $$= 50000 \lim_{t \to \infty} (1 - e^{-0.1t})$$

   $$= 50000(1 - 0) = 50000 \text{ vaccine units}$$

   So, the maximum production capacity is 50,000 vaccine units.

2. Solution:

   $$\lim_{t \to \infty} I(t) = \lim_{t \to \infty} \frac{100000t}{t + 50}$$

   $$= \lim_{t \to \infty} \frac{100000t}{t(1 + \frac{50}{t})} = \lim_{t \to \infty} \frac{100000}{1 + \frac{50}{t}}$$

   $$= \frac{100000}{1 + 0} = 100000 \text{ people}$$

   In the long term, information will reach a maximum of 100,000 people, indicating saturation in information spread.

3. Solution:

   Marginal cost is the derivative of the cost function, which can be calculated using the limit definition:

   $$\text{Marginal Cost} = \lim_{\Delta x \to 0} \frac{C(x + \Delta x) - C(x)}{\Delta x}$$

   Detailed steps:

   $$C(x + \Delta x) = 1000 + 5(x + \Delta x) + 0.01(x + \Delta x)^2$$

   We expand:

   $$(x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$$

   Then we substitute:

   $$C(x + \Delta x) = 1000 + 5x + 5\Delta x + 0.01x^2 + 0.02x\Delta x + 0.01(\Delta x)^2$$

   So,

   $$C(x + \Delta x) - C(x) = 5\Delta x + 0.02x\Delta x + 0.01(\Delta x)^2$$

   $$\frac{C(x + \Delta x) - C(x)}{\Delta x} = 5 + 0.02x + 0.01\Delta x$$

   $$\lim_{\Delta x \to 0} (5 + 0.02x + 0.01\Delta x) = 5 + 0.02x$$

   The marginal cost is $5 + 0.02x$ rupiah per unit. This means, to produce the $x$-th mask, the additional cost is $5 + 0.02x$ rupiah. The more production, the higher the marginal cost due to the term $0.02x$.