Limits in Context - Business Calculus

1. The number of active users on a new social media platform, in millions, is given by $U(t)$, where $t$ is the number of months since its launch. The expression $\frac{U(6) - U(3)}{3}$ represents the average rate of user growth over a specific period. Which limit precisely defines the instantaneous rate of user growth at the end of the 3rd month?
   Answer: $\lim_{h \to 0} \frac{U(3+h) - U(3)}{h}$ The instantaneous rate of change of a function $f(x)$ at a point $x=a$ is given by the definition of the derivative, $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$. To find the instantaneous rate of user growth at the end of the 3rd month ($t=3$), we must apply this definition to the function $U(t)$ at $a=3$. This yields the expression $\lim_{h \to 0} \frac{U(3+h) - U(3)}{h}$.
2. The demand function for a product is $q(p)$, where $q$ is the quantity demanded at price $p$. The elasticity of demand is given by $E(p) = \frac{-p \cdot q'(p)}{q(p)}$. If we know that for a certain price $p_0$, $\lim_{p \to p_0} q(p) = 500 \quad \text{and} \quad \lim_{h \to 0} \frac{q(p_0+h) - q(p_0)}{h} = -25$ What can be concluded about the demand at price $p_0$?
   Answer: The rate of change of demand with respect to price is $-25$, and the quantity demanded is approaching 500. When you encounter limits and derivatives together in a business calculus problem, you're being asked to interpret the fundamental definitions of these concepts in a real-world context. The given information tells us two key things through limit notation. First, $\lim_{p \to p_0} q(p) = 500$ means that as the price approaches $p_0$, the quantity demanded approaches 500 units. Second, $\lim_{h \to 0} \frac{q(p_0+h) - q(p_0)}{h} = -25$ is the definition of the derivative $q'(p_0)$, which represents the instantaneous rate of change of quantity with respect to price at $p_0$. Option D correctly interprets both pieces of information: the rate of change of demand with respect to price is -25 (meaning demand decreases by 25 units per dollar price increase), and quantity demanded approaches 500. Option A misses the negative sign, incorrectly stating demand is increasing when it's actually decreasing. Option B misinterprets the limit as a maximum constraint on total units and confuses the derivative value with a price, which makes no mathematical sense. Option C incorrectly assumes we can determine revenue behavior from quantity alone—revenue depends on both price and quantity, and we'd need the actual price value $p_0$ to calculate revenue. Remember that limits describe approaching values (not necessarily achieved values), and the derivative limit formula $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ always represents instantaneous rate of change. Pay close attention to signs—they indicate direction of change.
3. A manufacturing company's efficiency function is $E(n) = \frac{100n}{n + 25}$ percent, where $n$ is the number of experienced workers. The production manager finds that $\lim_{n \to 75} \frac{E(n) - E(75)}{n - 75} = 0.25$. How should this result guide staffing decisions?
   Answer: When the team has 75 experienced workers, efficiency is improving at a rate of 0.25 percentage points per additional worker When you encounter a limit expression like $\lim_{n \to 75} \frac{E(n) - E(75)}{n - 75}$, you're looking at the definition of a derivative at a specific point. This limit represents $E'(75)$, the instantaneous rate of change of efficiency when there are 75 experienced workers. The result of 0.25 tells us that at exactly 75 workers, efficiency is increasing at a rate of 0.25 percentage points per additional worker. This is what answer choice C correctly states - it's interpreting the derivative as a rate of change at that specific point. Let's examine why the other options miss the mark. Choice A misunderstands what the derivative represents - it gives the instantaneous rate, not the actual change from adding one worker. The derivative approximates change for small increments but isn't exact. Choice B incorrectly assumes this rate represents an optimal point. A positive derivative simply means efficiency is still increasing, but we'd need to find where $E'(n) = 0$ to locate a maximum. Choice D makes the critical error of assuming the rate of change remains constant beyond 75 workers. Derivatives typically vary as the input changes - this 0.25 rate applies specifically at $n = 75$. Remember that when you see the limit definition of a derivative in business contexts, it's asking about instantaneous rates of change at specific points, not overall optimization or constant rates across ranges. Focus on what the derivative means at that exact input value.
4. The cost function for a manufacturing process is $C(q) = 0.1q^2 + 5q + 200$ dollars, where $q$ is the quantity produced. An economist calculates $\lim_{q \to 50} \frac{C(q) - C(50)}{q - 50}$ to analyze production efficiency. What does this limit represent, and what is its value?
   Answer: Marginal cost at 50 units; equals $15.00 per additional unit produced This limit is the definition of the derivative C'(50), which represents marginal cost at q = 50. Taking the derivative: C'(q) = 0.2q + 5, so C'(50) = 0.2(50) + 5 = 15. Choice A confuses marginal cost with average cost. Choice C misinterprets the limit as a discrete change rather than an instantaneous rate. Choice D incorrectly identifies this as fixed cost, which would be the constant term.
5. A company's profit is $P(q)$ and its revenue is $R(q)$ for selling $q$ items. An analyst finds that $\lim_{q \to \infty} \frac{P(q)}{R(q)} = 0.12$ Assuming that revenue continues to grow as $q \to \infty$, what is the best interpretation of this limit?
   Answer: In the long run, the company's profit approaches 12% of its total revenue. The expression $\frac{P(q)}{R(q)}$ represents the profit margin, which is the ratio of profit to revenue. The limit describes the long-run behavior of this ratio. As the quantity produced and sold ($q$) becomes very large, this ratio approaches 0.12, or 12%. This means that for very large scales of operation, the company's profit tends to be about 12% of its total revenue.
6. A company is comparing two advertising campaigns. The number of new customers acquired from Campaign A is $N_A(t)$ and from Campaign B is $N_B(t)$, where $t$ is the number of days the campaign has run. At the two-week mark ($t=14$), their respective rates of acquisition are: $N'_A(14) = \lim_{h \to 0} \frac{N_A(14+h) - N_A(14)}{h} = 100$ $N'_B(14) = \lim_{h \to 0} \frac{N_B(14+h) - N_B(14)}{h} = 120$ What is the most valid conclusion that can be drawn from this data?
   Answer: On the 14th day, Campaign B is generating new customers at a higher rate than Campaign A. When you encounter derivatives in a business context, remember that the derivative represents the instantaneous rate of change at a specific point. The limit definition $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ tells you how fast something is changing at exactly time $t = a$. Here, $N'_A(14) = 100$ means Campaign A is acquiring new customers at a rate of 100 customers per day at the 14-day mark, while $N'_B(14) = 120$ means Campaign B is acquiring customers at 120 customers per day at that same moment. Since 120 > 100, Campaign B has a higher rate of customer acquisition on day 14, making D correct. Let's examine why the other options fail: A is wrong because derivatives tell us about rates, not cumulative totals. Campaign B could have a higher rate at day 14 but still have fewer total customers if it started poorly. B is incorrect because knowing the rate at one point doesn't predict long-term performance – rates can change over time. C makes a common error by confusing instantaneous rates with actual quantities. The derivative of 120 means the rate at day 14, not the number of customers gained on day 15. Study tip: Don't confuse derivatives (rates of change) with the original function values (actual quantities). When you see $f'(a)$, think "How fast is this changing at point $a$?" not "What's the total value?" This distinction appears frequently in business calculus applications.
7. An investment account's value follows $V(t) = 5000(1.08)^t$ dollars after $t$ years. An investor calculates $\lim_{t \to 10} \frac{V(t) - 10800}{t - 10} = 864$ to analyze the investment performance. What does this limit reveal about the investment at year 10?
   Answer: The account value is increasing at $864 per year when the balance reaches $10,800 at year 10 The limit represents V'(10), the instantaneous rate of change of account value at t = 10 years. Since V(10) = 5000$(1.08)^{10}$ ≈ 10,800, this means when the account reaches $10,800, it's growing at $864 per year at that instant. Choice B confuses instantaneous rate with actual earnings over a year. Choice C misinterprets this as average return rather than instantaneous rate. Choice D uses the instantaneous rate incorrectly to predict future value.
8. A population model for a city predicts $P(t) = \frac{80000}{1 + 15e^{-0.3t}}$ people after $t$ years. Urban planners need to understand the growth rate when the population reaches 40,000. If $\lim_{h \to 0} \frac{P(t_0 + h) - P(t_0)}{h} = 600$ when $P(t_0) = 40000$, what is the practical significance of this limit?
   Answer: At the moment when population is 40,000, it is growing at an instantaneous rate of 600 people per year The limit represents P'(t₀), the instantaneous rate of change of population at the specific time when P(t₀) = 40,000. This is the instantaneous growth rate of 600 people per year at that moment. Choice A confuses instantaneous rate with actual change over a year. Choice C misinterprets this as average rate over an interval. Choice D incorrectly describes this as acceleration (second derivative) rather than velocity (first derivative).
9. An online retailer's daily profit function is $P(x) = -0.02x^2 + 8x - 150$ dollars, where $x$ is the number of items sold. The function $\frac{P(x) - P(100)}{x - 100}$ represents the average rate of profit change. What happens to this average rate as $x$ approaches 100, and what is the business interpretation?
   Answer: The average rate approaches $4.00, representing the instantaneous rate of profit change when selling exactly 100 items As x approaches 100, the average rate of change approaches the instantaneous rate of change (derivative) at x = 100. P'(x) = -0.04x + 8, so P'(100) = 4. This represents the marginal profit at 100 items. Choice A confuses instantaneous rate with exact change per unit. Choice C misinterprets this as the profit from the 100th item rather than the rate of change. Choice D is incorrect because the limit exists and equals 4.
10. The concentration of a drug in the bloodstream follows $C(t) = \frac{12t}{t^2 + 4}$ mg/L, where $t$ is time in hours after injection. A pharmacologist needs to determine when the concentration is changing most rapidly. If $\lim_{h \to 0} \frac{C(2+h) - C(2)}{h} = 0$, what does this tell us about the drug concentration at $t = 2$ hours?
    Answer: The concentration is neither increasing nor decreasing at t = 2 hours, suggesting a critical point When you see a limit expression like $\lim_{h \to 0} \frac{C(2+h) - C(2)}{h}$, you're looking at the definition of a derivative at a specific point. This limit represents $C'(2)$, the instantaneous rate of change of drug concentration at $t = 2$ hours. Since this limit equals 0, we know that $C'(2) = 0$, meaning the rate of change of concentration is zero at $t = 2$. When the derivative equals zero, the function is neither increasing nor decreasing at that instant – it's at a critical point where the tangent line is horizontal. This suggests we've found either a maximum or minimum concentration. Option D correctly identifies this: the concentration is neither increasing nor decreasing at $t = 2$, indicating a critical point. Option A is wrong because knowing only that $C'(2) = 0$ doesn't tell us whether this is a minimum or maximum – we'd need the second derivative test or further analysis to determine that. Option B misinterprets what the limit represents. The instantaneous rate at $t = 2$ being zero has nothing to do with the average rate over the first 2 hours, which would be $\frac{C(2) - C(0)}{2 - 0}$. Option C makes an impossible claim. A derivative being zero at one point doesn't make the function constant afterward – the concentration will continue to change after $t = 2$. Remember: whenever you see the formal limit definition of a derivative equaling zero, you've found a critical point where the function's rate of change momentarily stops.
11. The profit, in thousands of dollars, from manufacturing $x$ hundred specialized microchips is given by the function $P(x)$. A calculation shows that $\lim_{h \to 0} \frac{P(20+h) - P(20)}{h} = 4.5$ Based on this result, which conclusion is the most appropriate?
    Answer: At a production level of 2,000 microchips, the marginal profit is $4,500 per hundred microchips. The limit represents the derivative $P'(20)$. Since $x$ is in hundreds of units, $x=20$ corresponds to $20 \times 100 = 2,000$ microchips. $P(x)$ is in thousands of dollars. The derivative $P'(x)$ is therefore in units of (thousands of dollars) per (hundred units). So, $P'(20) = 4.5$ means the marginal profit is $4.5 thousand dollars per hundred microchips, or $4,500 per hundred microchips, at the production level of 2,000 units.
12. A company's revenue function is $R(x) = 50x - 0.5x^2$ thousand dollars for $x$ thousand units sold. The marketing department reports that $\lim_{x \to 30^-} \frac{R(x) - R(30)}{x - 30} = 20$ and $\lim_{x \to 30^+} \frac{R(x) - R(30)}{x - 30} = 20$. What can be concluded about the marginal revenue at 30,000 units?
    Answer: Marginal revenue exists and equals $20,000 per thousand additional units, indicating revenue is still increasing Since both one-sided limits equal 20, the limit exists and equals 20, meaning the marginal revenue exists and equals $20,000 per thousand units (or $20 per unit). Since this is positive, revenue is still increasing at this production level. Choice B is wrong because both one-sided limits exist and are equal. Choice C has correct value but incorrect claim about maximum. Choice D makes an unsupported claim about optimality and incorrectly relates marginal to average revenue.
13. The value of an investment, $V(t)$, in thousands of dollars, is modeled as a function of time $t$ in years. The function is continuous for $t > 0$. However, due to a sudden market crash at $t=3$, the company must re-evaluate the asset. The value is modeled such that $\lim_{t \to 3} V(t) = 50$, but the officially recorded value is $V(3) = 35$. What is the best financial interpretation of this situation?
    Answer: Just before and just after 3 years, the investment's value was trending toward $50,000, but an immediate event caused a sudden write-down of its value on that day. The statement $\lim_{t \to 3} V(t) = 50$ means that as time approached 3 years from both the left and the right, the value of the investment was heading towards $50,000. However, the actual function value $V(3) = 35$ indicates that at the precise moment $t=3$, the value was different. This represents a removable discontinuity, best explained by a sudden, instantaneous event (like a market crash and subsequent accounting adjustment) that changed the asset's value from its expected trend.
14. A technology startup's user base grows according to $U(t) = \frac{50000t}{t + 10}$ users, where $t$ is months since launch. The growth rate at month 20 is calculated as $\lim_{\Delta t \to 0} \frac{U(20 + \Delta t) - U(20)}{\Delta t}$. If this limit equals 555.56, what strategic insight does this provide?
    Answer: At month 20, the user base is growing at approximately 556 new users per month instantaneously When you encounter a limit expression like $\lim_{\Delta t \to 0} \frac{U(20 + \Delta t) - U(20)}{\Delta t}$, you're looking at the definition of a derivative. This measures the instantaneous rate of change of the function $U(t)$ at $t = 20$. In business contexts, this tells you how fast something is changing at a specific moment in time. The given limit equals 555.56, which means $U'(20) = 555.56$. Since $U(t)$ represents users and $t$ represents months, the derivative $U'(20)$ has units of users per month. This is the instantaneous growth rate of the user base at month 20 — essentially a "snapshot" of how quickly users are being added at that precise moment. Answer C correctly interprets this as the instantaneous growth rate of approximately 556 new users per month at month 20. Answer A confuses the derivative with actual user count and incorrectly adds the growth rate to predict future users. Answer B misinterprets the instantaneous rate as a sustained average rate — the derivative tells you the rate at one moment, not a long-term average. Answer D completely misreads the context, introducing cost information that doesn't exist in the problem. Remember: when you see the limit definition of a derivative in business problems, you're always dealing with instantaneous rates of change. Don't confuse this with total amounts, averages over time, or unrelated business metrics. The derivative gives you the "speedometer reading" at one specific point.
15. The cost $C(x)$, in thousands of dollars, to remove $x$ percent of a toxic substance from a water supply is given by a function, where $0 \le x < 100$. It is determined that $\lim_{x \to 100^-} C(x) = +\infty$ What does this limit imply about the cleanup operation?
    Answer: Removing 100% of the substance is not just technologically difficult, but the cost to do so becomes prohibitively large. The limit indicates that as the percentage of the substance removed, $x$, approaches 100% from the left side (since you can't remove more than 100%), the cost $C(x)$ increases without bound. This represents a vertical asymptote at $x=100$. In a business or economic context, this means the cost would become infinitely large, making it practically impossible to achieve 100% removal due to prohibitive costs.
16. A company's profit function is $P(x) = -2x^3 + 15x^2 + 36x - 50$ thousand dollars, where $x$ is the number of units produced (in hundreds). If $\lim_{h \to 0} \frac{P(4+h) - P(4)}{h} = 12$, what is the most accurate interpretation of this limit in the business context?
    Answer: When producing 400 units, profit is increasing at a rate of $12,000 per hundred additional units produced The limit represents the derivative P'(4) = 12, which is the instantaneous rate of change of profit with respect to x at x = 4. Since x is measured in hundreds of units and P(x) is in thousands of dollars, the rate is $12,000 per hundred additional units. Choice B has wrong units (should be per hundred units, not per unit). Choice C confuses instantaneous rate with total change over an interval. Choice D incorrectly suggests this rate indicates optimality.
17. Let $\bar{C}(q) = \frac{C(q)}{q}$ be the average cost function for manufacturing $q$ electronic components. A financial analyst determines that $\lim_{q \to \infty} \bar{C}(q) = 15$ What is the most accurate business implication of this finding?
    Answer: As the number of components produced becomes very large, the average cost per component approaches $15. The limit $\lim_{q \to \infty} \bar{C}(q) = 15$ describes the long-run behavior of the average cost function. It means that as the production quantity $q$ increases without bound, the average cost per unit gets closer and closer to $15. This is often due to fixed costs being spread over a very large number of units.
18. Let $R(p)$ be the daily revenue in dollars from selling a smartphone app at a price of $p$ dollars. An analysis of the revenue function yields: $\lim_{p \to 9.99} \frac{R(p) - R(9.99)}{p - 9.99} = -500$ What is the correct interpretation of this limit?
    Answer: When the price is set at $9.99, the revenue is decreasing at a rate of $500 per dollar increase in price. The limit expression is the definition of the derivative, $R'(p)$, at $p=9.99$. The value of the derivative, -500, represents the instantaneous rate of change of revenue with respect to price. The negative sign indicates that revenue is decreasing. Therefore, when the price is $9.99, revenue is decreasing at a rate of $500 for each one-dollar increase in the price $p$.
19. The total shipping cost $S(w)$ for a parcel of weight $w$ pounds is modeled by a function. The function has a discontinuity at $w=50$, described by: $\lim_{w \to 50^-} S(w) = 40 \quad \text{and} \quad \lim_{w \to 50^+} S(w) = 35$ Which business practice best explains this discontinuity?
    Answer: A bulk-shipping discount takes effect for parcels weighing 50 pounds or more. The limits describe a jump discontinuity. As the weight $w$ approaches 50 pounds from below, the shipping cost approaches $40. As $w$ approaches 50 pounds from above, the cost approaches $35. The cost abruptly drops at the 50-pound mark. This is characteristic of a bulk discount where crossing a certain threshold (50 pounds) results in a lower price structure, possibly due to a flat rate or lower per-pound rate that makes the total cost jump down.
20. The total cost, in dollars, for a company to produce $q$ units of a product is given by the function $C(q)$. The following limit is known: $\lim_{h \to 0} \frac{C(500+h) - C(500)}{h} = 32$ What is the best interpretation of this mathematical statement in a business context?
    Answer: When 500 units are being produced, the cost is increasing at a rate of $32 per unit. The expression $\lim_{h \to 0} \frac{C(500+h) - C(500)}{h}$ is the definition of the derivative of the cost function, $C'(q)$, evaluated at $q=500$. This represents the instantaneous rate of change of cost with respect to the number of units produced. Therefore, $C'(500) = 32$ means that when production is at 500 units, the cost is increasing at a rate of $32 per unit. This value is also known as the marginal cost, which approximates the cost of producing the 501st unit.

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