Limits at Infinity - Business Calculus

1. A function is given by $f(x) = \frac{6e^{4x} - 10e^{-2x}}{2e^{4x} + 5e^{-2x}}$. The graph of this function has two distinct horizontal asymptotes. What is the sum of the $y$-values of these two asymptotes?
   Answer: $1$ We need to find the limits as $x \to \infty$ and $x \to -\infty$. For $x \to \infty$, the dominant term is $e^{4x}$. Divide the numerator and denominator by $e^{4x}$: $\lim_{x \to \infty} \frac{6 - 10e^{-6x}}{2 + 5e^{-6x}} = \frac{6 - 0}{2 + 0} = 3$ So, $y=3$ is one horizontal asymptote. For $x \to -\infty$, the dominant term is $e^{-2x}$ (since its exponent becomes positive). Divide the numerator and denominator by $e^{-2x}$: $\lim_{x \to -\infty} \frac{6e^{6x} - 10}{2e^{6x} + 5} = \frac{0 - 10}{0 + 5} = -2$ So, $y=-2$ is the other horizontal asymptote. The sum of the $y$-values is $3 + (-2) = 1$. Distractors B and C are the individual asymptote values. Distractor D is the difference between the values ($3 - (-2) = 5$).
2. According to this model, what is the company's projected market share in the long run?
   Answer: $60\%$ First, use the initial condition to find the constant $A$. We are given $S(0) = 15$. Plugging $t=0$ into the model gives: $S(0) = \frac{A(0+2)}{0+8} = \frac{2A}{8} = \frac{A}{4}$. Setting this equal to the given initial value: $\frac{A}{4} = 15$, which implies $A = 60$. Now, the model is $S(t) = \frac{60(t+2)}{t+8} = \frac{60t+120}{t+8}$. The long-run market share is the limit as $t \to \infty$: $\lim_{t \to \infty} S(t) = \lim_{t \to \infty} \frac{60t+120}{t+8}$ Since the degrees of the numerator and denominator are equal, the limit is the ratio of the leading coefficients, which is $\frac{60}{1} = 60$. So the projected market share is 60%. Distractor A is the initial share. Distractor D is a miscalculation, perhaps solving for $A/2$. Distractor C is the constant term ratio $120/1$, which is not how this limit is found.
3. A manufacturing company's efficiency ratio is given by $E(n) = \frac{6n^2 - 4n + 1}{3n^2 + 5n - 2}$ where $n$ is the number of employees (in hundreds). Management wants to know if doubling the workforce indefinitely will result in an efficiency ratio greater than 1.8. Based on the limit analysis, what should management conclude?
   Answer: Doubling the workforce will result in efficiency approaching exactly 2, which exceeds the 1.8 target threshold When analyzing how a rational function behaves as the input grows very large, you need to examine the horizontal asymptote by comparing the degrees of the numerator and denominator polynomials. For $E(n) = \frac{6n^2 - 4n + 1}{3n^2 + 5n - 2}$, both the numerator and denominator are degree 2 polynomials. When the degrees are equal, the horizontal asymptote equals the ratio of the leading coefficients. Here, that's $\frac{6}{3} = 2$. This means as $n \to \infty$ (representing indefinitely doubling the workforce), $E(n) \to 2$. Since 2 > 1.8, the efficiency will approach a value that exceeds management's target threshold. Let's examine why the other answers are wrong: A is incorrect because efficiency doesn't increase without bound—it approaches the finite limit of 2, not infinity. B contains a calculation error. The horizontal asymptote is $\frac{6}{3} = 2$, not 1.5. Perhaps this represents confusion about which coefficients to use or an arithmetic mistake. C is completely wrong about the limiting behavior. When both polynomials have the same degree, the limit is never zero—that only happens when the denominator's degree exceeds the numerator's degree. D correctly identifies that efficiency approaches exactly 2, which does exceed the 1.8 target. Study tip: For rational functions where numerator and denominator have equal degrees, always find the horizontal asymptote by dividing the leading coefficients. This gives you the long-term behavior that's crucial for business planning decisions.
4. Find the limit: $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x/2}$
   Answer: $e^{3/2}$ When you encounter a limit of the form $(1 + \text{something})^{\text{something else}}$ as the variable approaches infinity, you're dealing with a variation of the fundamental exponential limit that defines Euler's number $e$. The key insight is recognizing this pattern and transforming it into the standard form. The standard limit is $\lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^u = e$. To solve your problem, you need to manipulate the expression to match this form. Starting with $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{x/2}$, let's substitute $u = \frac{x}{3}$, which means $x = 3u$ and as $x \to \infty$, we have $u \to \infty$. The expression becomes: $\lim_{u \to \infty} \left(1 + \frac{3}{3u}\right)^{3u/2} = \lim_{u \to \infty} \left(1 + \frac{1}{u}\right)^{3u/2}$ This can be rewritten as: $\lim_{u \to \infty} \left[\left(1 + \frac{1}{u}\right)^u\right]^{3/2} = e^{3/2}$ Looking at the wrong answers: (A) $e^6$ results from incorrectly multiplying the coefficient and exponent ($3 \times 2 = 6$). (B) $1$ ignores the exponential growth entirely. (C) $e^{2/3}$ flips the fraction, possibly from confusing the roles of the numerator and denominator in the transformation. Strategy tip: For limits involving $(1 + \text{something})^{\text{something else}}$, always look for ways to transform them into the standard form $\left(1 + \frac{1}{u}\right)^u = e$ through substitution. The final exponent in your answer will be the coefficient that emerges from this transformation.
5. Consider the function $f(x) = \frac{5x^3 + 2x - 7}{x^4 - x^2 + 3}$. A business analyst needs to determine the end behavior of this function to model long-term trends. Which of the following correctly describes both horizontal asymptotes?
   Answer: There is a horizontal asymptote at $y = 0$ approached from both directions as $x \to \pm\infty$ Since the degree of the denominator (4) is greater than the degree of the numerator (3), the limit as x approaches ±∞ is 0. This creates a horizontal asymptote at y = 0. Choice B incorrectly uses the leading coefficient of the numerator. Choice C incorrectly assumes different behavior for positive and negative infinity, which doesn't occur here. Choice D incorrectly suggests no horizontal asymptotes exist when the denominator has higher degree.
6. Consider the piecewise function $f(x)$ defined as: $f(x) = \begin{cases} \frac{10x^3 - 7x}{2x^3 + 5} & \text{if } x \ge 0 \\ \frac{15x-2}{\sqrt{9x^2+1}} & \text{if } x < 0 \end{cases}$ Which of the following statements correctly describes the horizontal asymptotes of $f(x)$?
   Answer: The function has two horizontal asymptotes, at $y=5$ and $y=-5$. When analyzing horizontal asymptotes for piecewise functions, you need to examine the behavior as x approaches both positive and negative infinity, checking each piece of the function separately. For the first piece (when x ≥ 0), as x → +∞, you have $\frac{10x^3 - 7x}{2x^3 + 5}$. To find this limit, divide both numerator and denominator by the highest power of x, which is x³: $\lim_{x \to +\infty} \frac{10 - \frac{7}{x^2}}{2 + \frac{5}{x^3}} = \frac{10}{2} = 5$ For the second piece (when x < 0), as x → -∞, you have $\frac{15x-2}{\sqrt{9x^2+1}}$. Since x is negative, factor out |x| = -x from under the square root: $\sqrt{9x^2+1} = |x|\sqrt{9+\frac{1}{x^2}} = -x\sqrt{9+\frac{1}{x^2}}$ (since x < 0). This gives: $\lim_{x \to -\infty} \frac{15x-2}{-x\sqrt{9+\frac{1}{x^2}}} = \lim_{x \to -\infty} \frac{15-\frac{2}{x}}{-\sqrt{9+\frac{1}{x^2}}} = \frac{15}{-3} = -5$ Choice A identifies only one asymptote, missing the left-side behavior. Choice B incorrectly calculates the left asymptote as -15. Choice C makes an error by suggesting 15/9 instead of the correct value. Choice D correctly identifies both horizontal asymptotes: y = 5 (as x → +∞) and y = -5 (as x → -∞). Remember: piecewise functions can have different horizontal asymptotes on each side, so always check both directions separately and pay careful attention to signs when dealing with square roots.
7. A function is defined by $f(x) = \frac{(kx-3)(2x+5)}{6x^2 - 7x + 1}$. For what value of the constant $k$ does the graph of $f(x)$ have a horizontal asymptote at $y=2$?
   Answer: $k = 6$ A horizontal asymptote is determined by the limit of the function as $x \to \infty$ or $x \to -\infty$. For a rational function where the degrees of the numerator and denominator are equal, this limit is the ratio of the leading coefficients. First, expand the numerator: $(kx-3)(2x+5) = 2kx^2 + 5kx - 6x - 15 = 2kx^2 + (5k-6)x - 15$. The leading term of the numerator is $2kx^2$ and its coefficient is $2k$. The leading term of the denominator is $6x^2$ and its coefficient is $6$. The limit is $\lim_{x \to \infty} f(x) = \frac{2k}{6} = \frac{k}{3}$. To have a horizontal asymptote at $y=2$, we set the limit equal to 2: $\frac{k}{3} = 2$, which gives $k=6$.
8. A logistics company models delivery time efficiency using $T(d) = 5 - \frac{3d^2 + 2d}{d^3 + d + 7}$ where $d$ represents distance in hundreds of miles. For very long delivery routes, what will happen to the delivery time efficiency?
   Answer: The efficiency approaches exactly 5 units as distance increases indefinitely When you encounter questions about the long-term behavior of business functions, you're dealing with limits at infinity. This concept helps predict what happens to key metrics like efficiency, cost, or revenue as variables grow very large. To find what happens to $T(d)$ as distance increases indefinitely, you need to evaluate $\lim_{d \to \infty} T(d) = \lim_{d \to \infty} \left(5 - \frac{3d^2 + 2d}{d^3 + d + 7}\right)$. For the fraction $\frac{3d^2 + 2d}{d^3 + d + 7}$, divide both numerator and denominator by the highest power of $d$ in the denominator (which is $d^3$): $\frac{3d^2 + 2d}{d^3 + d + 7} = \frac{\frac{3}{d} + \frac{2}{d^2}}{1 + \frac{1}{d^2} + \frac{7}{d^3}}$ As $d \to \infty$, all terms with $d$ in the denominator approach zero, so this fraction approaches $\frac{0}{1} = 0$. Therefore, $T(d)$ approaches $5 - 0 = 5$. Answer A is incorrect because it gives 2, which would result from mishandling the algebraic manipulation. Answer C is wrong because 3 might come from focusing only on the leading coefficient of the numerator while ignoring the subtraction structure. Answer D is incorrect because the function actually approaches a finite limit rather than decreasing without bound. Remember: when analyzing rational functions at infinity, always divide by the highest power in the denominator. The dominant terms determine the limit, while lower-order terms vanish.
9. What is the value of the limit $L = \lim_{x \to \infty} \frac{\ln(e^{5x} + x^2)}{3x + \ln(x)}$?
   Answer: $L = \frac{5}{3}$ For large $x$, the term $e^{5x}$ dominates $x^2$ in the argument of the natural logarithm, and $3x$ dominates $\ln(x)$ in the denominator. Therefore, we can approximate the limit by considering the dominant terms: $L = \lim_{x \to \infty} \frac{\ln(e^{5x})}{3x}$ Using the logarithm property $\ln(a^b)=b\ln(a)$, we get: $L = \lim_{x \to \infty} \frac{5x}{3x} = \frac{5}{3}$ Alternatively, one could use L'Hôpital's Rule. The derivative of the numerator is $\frac{5e^{5x}+2x}{e^{5x}+x^2}$. The derivative of the denominator is $3 + \frac{1}{x}$. The limit of their ratio is $\lim_{x \to \infty} \frac{(5e^{5x}+2x)/(e^{5x}+x^2)}{(3+1/x)} = \lim_{x \to \infty} \frac{5+2x/e^{5x}}{1+x^2/e^{5x}} \cdot \frac{1}{3+1/x} = \frac{5+0}{1+0} \cdot \frac{1}{3+0} = \frac{5}{3}$. Distractor B might come from focusing on the $x^2$ term, getting $2\ln(x)$, and then taking a ratio with $3x$, which is incorrect. Distractor C might come from adding the coefficients in the denominator: $3x+\ln(x)$ -> $3+1=4$.
10. What is the long-run average cost per hundred units?
    Answer: $200,000 The average cost function $\bar{C}(x)$ is the total cost $C(x)$ divided by the number of units $x$. Here, units are in hundreds. So, $\bar{C}(x) = \frac{C(x)}{x} = \frac{200x^2 + 1500}{x(x+10)} = \frac{200x^2 + 1500}{x^2 + 10x}$. The long-run average cost is the limit of $\bar{C}(x)$ as $x \to \infty$. Since the degrees of the numerator and denominator are both 2, the limit is the ratio of the leading coefficients: $\lim_{x \to \infty} \bar{C}(x) = \lim_{x \to \infty} \frac{200x^2 + 1500}{x^2 + 10x} = \frac{200}{1} = 200$ Since the cost $C(x)$ is in thousands of dollars, the long-run average cost is $200$ thousand dollars, or $200,000. Distractor D incorrectly analyzes the limit of $C(x)$, not $\bar{C}(x)$. Distractor B comes from taking the ratio of the constant terms, $\frac{1500}{10}=150$, which is an incorrect method. Distractor C is a miscalculation.
11. An environmental consultant models pollution reduction effectiveness using $P(t) = \frac{4t^3 + t^2 - 2}{t^4 + 2t^3 + 1} + 0.8$ where $t$ represents time in years after implementing new regulations. For very long-term environmental planning, what should be expected about pollution reduction effectiveness?
    Answer: The effectiveness will approach exactly 0.8 units in the long term When you encounter a function asking about long-term behavior, you need to find the limit as time approaches infinity. This tests your understanding of how rational functions behave over extended periods. To find what happens to $P(t) = \frac{4t^3 + t^2 - 2}{t^4 + 2t^3 + 1} + 0.8$ as $t \to \infty$, focus on the rational function part first. For rational functions, the long-term behavior is determined by the highest-degree terms in the numerator and denominator. The numerator's highest term is $4t^3$ and the denominator's is $t^4$, so $\frac{4t^3}{t^4} = \frac{4}{t}$. As $t \to \infty$, this fraction approaches 0. Therefore, the entire rational function approaches 0, and $P(t)$ approaches $0 + 0.8 = 0.8$. Choice A (4 units) incorrectly uses the coefficient of the numerator's highest-degree term while ignoring that the denominator has higher degree. Choice C (4.8 units) makes the same error but adds the constant 0.8. This represents the common misconception of adding coefficients rather than properly evaluating the limit. Choice D (0 units) correctly identifies that the rational part goes to zero but forgets about the constant term $+0.8$. Remember: when finding limits at infinity for rational functions, compare the degrees of numerator and denominator first. If the denominator has higher degree, the limit is zero. Always account for any constants added or subtracted from the rational expression.
12. Evaluate the limit: $\lim_{x \to -\infty} \frac{5x - 4}{\sqrt{9x^2 + 2x - 1}}$
    Answer: $-\frac{5}{3}$ To evaluate the limit as $x \to -\infty$, we divide both numerator and denominator by $|x|$. Since $x < 0$, we have $|x| = -x$. For the numerator: $\frac{5x-4}{-x} = \frac{5x-4}{-x} = -5 + \frac{4}{x}$. For the denominator: $\frac{\sqrt{9x^2 + 2x - 1}}{-x} = \frac{\sqrt{9x^2 + 2x - 1}}{\sqrt{x^2}} = \sqrt{\frac{9x^2+2x-1}{x^2}} = \sqrt{9 + \frac{2}{x} - \frac{1}{x^2}}$. The limit becomes: $\lim_{x \to -\infty} \frac{-5 + \frac{4}{x}}{\sqrt{9 + \frac{2}{x} - \frac{1}{x^2}}} = \frac{-5 + 0}{\sqrt{9 + 0 - 0}} = \frac{-5}{3} = -\frac{5}{3}$
13. An economist models the long-term unemployment rate using $U(t) = 3 + \frac{2t + 5}{t^2 + 4t + 1}$ where $t$ represents years after implementation of a new policy. After analyzing this model, what can be concluded about the unemployment rate's long-term behavior?
    Answer: The unemployment rate will stabilize at exactly 3% as the policy effects mature over many years When you encounter a function modeling long-term behavior, you need to find the limit as time approaches infinity. This tells you what value the function approaches after many years have passed. To find the long-term unemployment rate, calculate $\lim_{t \to \infty} U(t) = \lim_{t \to \infty} \left(3 + \frac{2t + 5}{t^2 + 4t + 1}\right)$. The key insight is analyzing what happens to the fraction $\frac{2t + 5}{t^2 + 4t + 1}$ as $t$ gets very large. Since the denominator has the highest power of $t$ (specifically $t^2$), it grows much faster than the numerator (which has $t^1$). When the denominator of a fraction grows faster than the numerator, the entire fraction approaches zero. More formally, dividing both numerator and denominator by $t^2$ gives: $\frac{2t + 5}{t^2 + 4t + 1} = \frac{\frac{2}{t} + \frac{5}{t^2}}{1 + \frac{4}{t} + \frac{1}{t^2}}$. As $t \to \infty$, all terms with $t$ in the denominator approach zero, leaving $\frac{0}{1} = 0$. Therefore, $\lim_{t \to \infty} U(t) = 3 + 0 = 3$, confirming answer D. Answer A incorrectly suggests the entire function approaches zero. Answer B (2%) likely results from mistakenly using just the coefficient of $t$ in the numerator. Answer C (5%) probably comes from using only the constant term in the numerator, ignoring the limit process entirely. Study tip: For rational functions, the long-term behavior depends on the highest powers in the numerator and denominator. When the denominator's degree exceeds the numerator's, that portion always approaches zero.
14. What is the horizontal asymptote of the function $g(x) = \frac{(2x-1)^3}{3x(2x+1)^2}$?
    Answer: $y = \frac{2}{3}$ When finding horizontal asymptotes of rational functions, you need to compare the degrees of the numerator and denominator after expanding to standard polynomial form. Let's expand this function. The numerator $(2x-1)^3$ has degree 3, and when expanded, its leading term is $(2x)^3 = 8x^3$. The denominator $3x(2x+1)^2$ also has degree 3. The term $(2x+1)^2$ expands to $(2x)^2 + \text{lower terms} = 4x^2 + \text{lower terms}$, so the denominator becomes $3x \cdot 4x^2 + \text{lower terms} = 12x^3 + \text{lower terms}$. Since both numerator and denominator have the same degree (3), the horizontal asymptote equals the ratio of their leading coefficients: $y = \frac{8}{12} = \frac{2}{3}$ Looking at the wrong answers: Choice A gives $y = \frac{8}{12}$ but leaves the fraction unreduced—while mathematically equivalent to $\frac{2}{3}$, it's not in simplest form. Choice B suggests $y = 2$, which might come from incorrectly thinking the leading coefficient of the numerator is 2 and ignoring the denominator's leading coefficient entirely. Choice D gives $y = 0$, which would be correct if the numerator's degree were less than the denominator's degree, but both degrees are equal here. Study tip: For rational functions, always expand to find the true degrees and leading coefficients. When degrees are equal, the horizontal asymptote is always the ratio of leading coefficients, simplified to lowest terms.
15. A retail chain's revenue per store is modeled by $R(s) = \frac{8s^3 - 3s^2 + s}{2s^3 + s^2 - 4}$ where $s$ is the number of stores in thousands. The corporate board needs to determine if opening stores indefinitely will eventually yield a revenue per store ratio exceeding 3.5. What analysis should guide their decision?
    Answer: The limit analysis shows revenue per store approaches 4, indicating the expansion strategy will succeed When you encounter a business problem asking about long-term behavior "as stores increase indefinitely," you're dealing with limits at infinity. The key insight is that rational functions approach a horizontal asymptote that depends on the degrees and leading coefficients of the numerator and denominator. To find $\lim_{s \to \infty} R(s) = \lim_{s \to \infty} \frac{8s^3 - 3s^2 + s}{2s^3 + s^2 - 4}$, focus on the highest-degree terms. Both numerator and denominator are degree 3, so divide both by $s^3$: $\lim_{s \to \infty} \frac{8 - \frac{3}{s} + \frac{1}{s^2}}{2 + \frac{1}{s} - \frac{4}{s^3}}$ As $s \to \infty$, all terms with $s$ in the denominator approach zero, leaving $\frac{8}{2} = 4$. Since 4 > 3.5, the expansion strategy will succeed. Answer A incorrectly uses 8 instead of the ratio $\frac{8}{2}$ - this represents focusing only on the numerator's leading coefficient. Answer B gives 3, which might result from computational errors or incorrectly handling the leading coefficients. Answer D gives 2, possibly from using the denominator's leading coefficient alone or making sign errors. Remember: For rational functions where numerator and denominator have equal degrees, the horizontal asymptote equals the ratio of the leading coefficients. This technique is crucial for analyzing long-term business trends modeled by rational functions.
16. A company's quarterly profit function is given by $P(x) = \frac{3x^3 - 2x^2 + 5x}{x^3 + x + 1}$ where $x$ represents the number of quarters since the company started operations. As the company operates for many quarters, what can be concluded about the long-term behavior of quarterly profits?
    Answer: The quarterly profit approaches $3$ as a horizontal asymptote, indicating stable long-term profitability To find the limit as x approaches infinity, we examine the highest degree terms in the numerator and denominator. Both have degree 3, so we take the ratio of their coefficients: lim(x→∞) P(x) = 3/1 = 3. This means there's a horizontal asymptote at y = 3. Choice B incorrectly assumes the limit is 0 (which would occur if the denominator had higher degree). Choice C ignores that rational functions with equal degrees in numerator and denominator have horizontal asymptotes. Choice D uses the coefficient of the x² term instead of the x³ term.
17. Find the horizontal asymptote of the function $f(x) = \sqrt{9x^2+12x} - 3x$.
    Answer: $y=2$ When you encounter a function involving square roots and polynomial terms like this, you're dealing with a horizontal asymptote problem that requires algebraic manipulation. The key insight is recognizing that direct substitution of large values won't work due to the indeterminate form, so you need to rationalize. To find the horizontal asymptote of $f(x) = \sqrt{9x^2+12x} - 3x$, multiply by the conjugate: $\frac{\sqrt{9x^2+12x} + 3x}{\sqrt{9x^2+12x} + 3x}$ This gives you: $f(x) = \frac{(\sqrt{9x^2+12x})^2 - (3x)^2}{\sqrt{9x^2+12x} + 3x} = \frac{9x^2+12x - 9x^2}{\sqrt{9x^2+12x} + 3x} = \frac{12x}{\sqrt{9x^2+12x} + 3x}$ For large positive $x$, factor out $x$ from the square root: $\sqrt{9x^2+12x} = x\sqrt{9+\frac{12}{x}} \approx 3x$ as $x \to \infty$ Therefore: $\lim_{x \to \infty} \frac{12x}{3x + 3x} = \frac{12x}{6x} = 2$ The horizontal asymptote is $y = 2$, making D correct. Choice A ($y=0$) would result from incorrectly assuming the denominator grows faster than the numerator. Choice B (no asymptote) ignores that rational functions typically have horizontal asymptotes when degrees are equal. Choice C ($y=4$) likely comes from miscalculating the limit or making an algebraic error. Study tip: For functions involving $\sqrt{ax^2+bx} - cx$, always rationalize by multiplying by the conjugate. This converts the problem into a standard rational function limit that's much easier to evaluate.
18. A pharmaceutical company's research budget allocation ratio is given by $B(y) = \frac{2y^4 - y^3 + 3y}{y^4 + 5y^2 - 1}$ where $y$ represents years of operation. The board wants to determine if this ratio will eventually stabilize above 1.5 for long-term planning purposes. What conclusion should they reach?
    Answer: The ratio will stabilize at exactly 2, which meets their long-term planning requirement Since both numerator and denominator have degree 4, we take the ratio of leading coefficients: lim(y→∞) B(y) = 2/1 = 2. Since 2 > 1.5, their planning requirement is met. Choice B incorrectly calculates the limit, possibly using coefficients from lower-degree terms. Choice C incorrectly calculates the limit, possibly using an incorrect ratio of coefficients. Choice D incorrectly assumes the function increases without bound when both polynomials have the same degree.
19. Consider the cost function $C(x) = \frac{x^2 + 3x + 2}{x + 1}$ where $x$ represents production volume in thousands of units. A financial analyst claims that as production increases indefinitely, the average cost per unit will approach a finite value. How should this claim be evaluated?
    Answer: The claim is incorrect because the cost function increases without bound as production increases indefinitely When you encounter questions about cost functions and long-term behavior, you need to analyze what happens as production volume approaches infinity. The key insight here is recognizing that this isn't about finding horizontal asymptotes—it's about understanding what the cost function actually represents. The given function $C(x) = \frac{x^2 + 3x + 2}{x + 1}$ represents total cost, not average cost per unit. To evaluate the analyst's claim about average cost, you need to find $\frac{C(x)}{x}$, which gives the average cost per unit. Let's simplify the original function first. Using polynomial long division: $C(x) = \frac{x^2 + 3x + 2}{x + 1} = x + 2 + \frac{0}{x + 1} = x + 2$ This shows that total cost grows linearly as $C(x) = x + 2$. Therefore, average cost per unit is $\frac{C(x)}{x} = \frac{x + 2}{x} = 1 + \frac{2}{x}$. As $x \to \infty$, this approaches 1, meaning average cost does approach a finite value. However, the question asks about the cost function itself, not average cost. Since $C(x) = x + 2$, the total cost increases without bound as production increases indefinitely, making answer C correct. Answer A incorrectly finds a horizontal asymptote at $y = 3$. Answer B incorrectly claims the asymptote is $y = 1$. Answer D wrongly suggests costs approach zero. Always distinguish between total cost and average cost in business problems—they behave very differently as production scales up.
20. The market share of a technology startup is modeled by $S(t) = \frac{4t^2 + 6t - 1}{2t^2 + 3t + 7}$ where $t$ is the number of years since launch. If a competitor claims that this startup's market share will eventually exceed 2.5%, which statement best evaluates this claim?
    Answer: The claim is false because the market share approaches exactly 2% as time increases indefinitely To find lim(t→∞) S(t), we look at the ratio of the leading coefficients since both numerator and denominator have degree 2: lim(t→∞) S(t) = 4/2 = 2. The horizontal asymptote is at y = 2, meaning the market share approaches 2%, which is less than 2.5%. Choice B incorrectly calculates the limit (possibly adding coefficients instead of dividing). Choice C incorrectly assumes the function decreases without bound. Choice D incorrectly assumes the function increases without bound.

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