Applications of Limits

A Simplified Public Health Model

Limits can describe the long-term behavior of a mathematical model. In public health, a formula must be fitted to reliable data and validated before anyone uses it to guide a real decision. The formula below shows how to calculate a limit and check the valid domain of a model.

Inspecting the Domain of the Model

This deliberately simplified formula is used to check how far the model stays meaningful:

\begin{aligned} &N(t) \\ &= 285000 - \sqrt{t^2 - t + (190.68)^3} \end{aligned}

Here, $N(t)$ is interpreted as a modeled case count, while $t$ is a nonnegative input whose unit must be defined for the scenario being studied.

To test whether the formula remains meaningful in the long run, calculate:

Abbreviate the square-root term as $R(t)$ :

R(t)=\sqrt{t^2-t+(190.68)^3}

Long-term behavior:

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 $t \gg 1$ ,

So,

For large positive $t$ :

\begin{aligned} &\sqrt{t^2 - t + (190.68)^3} \\ &\approx\sqrt{t^2}=|t|=t \end{aligned}

A negative case count means the formula has been used outside the model's valid domain. A case count cannot be negative, so the formula is meaningful only while $N(t) \geq 0$ . Read this simplified formula as a mathematical example of exponential growth inside the model, and keep it out of medical or policy decisions.

Model Output at a Vaccination Target

In this model, limits and function values show what the formula gives near a chosen vaccination target. That calculation does not establish that the target is effective. Evidence about the population and disease transmission is still required.

Suppose the vaccination target is $V \text{ people}$ among residents older than $18 \text{ years}$ . Substitute $V$ into the formula to calculate the case count given by the model at that target. This result comes from the formula alone and does not establish that the target is the best strategy.

Application in Economics

A marginal quantity measures the change produced by one extra unit. The limit turns that change into an instantaneous rate, which is what the marginal value means once the extra unit is made arbitrarily small.

Marginal Cost Analysis

In a differentiable cost model, marginal cost is the instantaneous rate at which total cost changes with output. It is defined by the derivative:

Mathematical definition using limits:

\begin{aligned} &\text{Marginal Cost} \\ &= \lim_{\Delta x \to 0} \frac{C(x + \Delta x) - C(x)}{\Delta x} \end{aligned}

The quantities in this definition are:

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

When output is counted in whole units, the exact additional cost of the next unit is $C(x+1)-C(x)$ . The derivative $C'(x)$ often approximates that discrete change, but the two quantities are not automatically identical.

Population Growth Model

Take positive parameters $K>0$ , $A>0$ , and $r>0$ . An idealized logistic population model then has the form:

P(t) = \frac{K}{1 + Ae^{-rt}}

Since $e^{-rt} \to 0$ as $t \to \infty$ , the model approaches its carrying-capacity parameter:

\lim_{t \to \infty} P(t) = \frac{K}{1+A(0)}=K

Here, $K$ is a parameter of the simplified model. A real environment's carrying capacity can change, so the limit does not claim that every measured population will settle permanently at one fixed value.

Application in Technology and Science

Each example applies a limit to a process that develops over time. In both cases the question is how one quantity develops while another quantity approaches a limit value.

Digital Signal Analysis

For a normalized first-order low-pass model with cutoff frequency $\omega_c > 0$ , the magnitude response is:

|H(i\omega)|=\frac{1}{\sqrt{1+(\omega/\omega_c)^2}}

Its limiting values make the idealized behavior explicit:

\lim_{\omega \to 0}|H(i\omega)|=1

This model passes very low frequencies with unit gain and increasingly attenuates high frequencies. It describes one kind of mathematical filter. Other digital filters can follow different rules.

Chemical Reaction Rate

For an idealized first-order consumption of reactant $A$ , the integrated rate law is:

[A](t)=[A]_0e^{-kt}, \qquad k>0

The calculation gives:

\lim_{t \to \infty}[A](t)=[A]_0\lim_{t \to \infty}e^{-kt}=0

This conclusion belongs to that irreversible first-order model. A reversible reaction at equilibrium generally requires a different model and need not have zero reactant concentration.

Calculating Cases at a Vaccination Target

Let $t$ be the number of vaccinated people. The following model calculates the number of remaining positive cases, $N(t)$ :

\begin{aligned} &N(t) \\ &= 285000 - \sqrt{t^2 - t + (190.68)^3} \end{aligned}

We will calculate $N(282{,}367)$ for a vaccination target of $282{,}367 \text{ people}$ .

Data and Assumptions

The calculation uses these assumptions:

Total population is $576{,}260 \text{ people}$ .
Children make up $21\%$ of the population and adults make up $79\%$ .
The positive-case proportion is $30\%$ in each age group.
Only adults outside the positive-case group are eligible for vaccination.

These assumptions give the following population counts:

\begin{aligned} \text{Children} &= 0.21 \times 576{,}260 \\ &= 121{,}014.6 \\ &\approx 121{,}015 \text{ people} \end{aligned}

Using the unrounded eligible count, the target represents:

\frac{282{,}367}{318{,}671.78}\times 100\% \approx 88.6\%

The target is about $88.6\%$ of the eligible adults in this calculation. This simplified formula only shows how to evaluate the function at that target. It cannot determine a real vaccination target.

Calculation

To calculate the model's case count, we use $t=282{,}367$ :

\begin{aligned} R_0 &=(282367)^2-282367 +(190.68)^3 \end{aligned}

Step 1 calculates the constant term.

\begin{aligned} &(190.68)^3 \\ &= 190.68 \times 190.68 \times 190.68 \\ &\approx 6{,}932{,}907.88 \end{aligned}

The displayed value is rounded to two decimal places. The next calculation retains the unrounded value internally.

Step 2 calculates the square of the vaccination count.

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

Step 3 combines the values under the square root.

\begin{aligned} &79731122689-282367 \\ &\quad{}+6932907.88 \\ &=79737773229.88 \end{aligned}

Step 4 gives the final result.

R_0\approx79737773229.88

Meaning of the Result

If $282{,}367$ people are vaccinated, the supplied formula gives approximately $2{,}621$ remaining cases. This result comes only from the simplified formula. Real data is still required for a forecast or a decision about health-resource allocation.

Limits of the Formula and Model

The value $2{,}621 \text{ cases}$ answers the mathematical question for the supplied formula. A real decision would still require independently validated evidence about:

Current and projected hospital capacity
Available medical personnel and workload
Resource allocation under several credible scenarios
Uncertainty and a truthful public communication strategy

A limit describes the behavior of a formula. Real data is still needed to determine whether the formula matches actual conditions. State the model's assumptions, valid domain, units, and uncertainty when explaining the calculated result.

Exercises

Each problem gives a situation and asks for the value a quantity approaches. Identify the quantity first, then evaluate the limit.

A pharmaceutical company models cumulative vaccine production with $P(t) = 50000(1 - e^{-0.1t})$ for $t \geq 0$ . Determine the long-term production ceiling predicted by the model.

Worked Solutions

Solution:

The model's long-term ceiling is obtained by calculating the limit when $t \to \infty$ :
$\lim_{t \to \infty} P(t) = \lim_{t \to \infty} 50000(1 - e^{-0.1t})$