One-Sided & Infinite Limits - Business Calculus

1. Let f(x) be a piecewise function defined as f(x) = \begin{cases} x-1, & \text{if } x \le 2 \\ 5-x, & \text{if } x > 2 \end{cases} and let g(x) = \frac{1}{x-2}. What is the value of \lim_{x \to 2^+} (f(x) \cdot g(x))?
   Answer: +\infty To evaluate the limit of the product, we can evaluate the limit of each function separately. For the limit x \to 2^+, we consider values of x slightly greater than 2. For f(x), we use the piece 5-x since x > 2. \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5-x) = 5 - 2 = 3. For g(x), as x \to 2^+, the denominator x-2 approaches 0 from the positive side (0^+). So, \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} \frac{1}{x-2} = +\infty. Now, we multiply the results: \lim_{x \to 2^+} (f(x) \cdot g(x)) = (3) \cdot (+\infty) = +\infty.
2. Consider the function $h(x) = \frac{\ln(x-1)}{x-2}$. What is the behavior of this function as $x$ approaches 2 from the left and from the right?
   Answer: Left-hand limit is $-\infty$ and right-hand limit is $+\infty$ due to denominator sign change As $x \to 2$, the numerator $\ln(x-1) \to \ln(1) = 0$ and denominator $(x-2) \to 0$. This is a $\frac{0}{0}$ form requiring L'Hôpital's rule or careful analysis. Using L'Hôpital's rule: $\lim_{x \to 2} \frac{\ln(x-1)}{x-2} = \lim_{x \to 2} \frac{\frac{1}{x-1}}{1} = \frac{1}{2-1} = 1$. However, this assumes the limit exists. For one-sided limits: as $x \to 2^-$, $\ln(x-1) \to 0^+$ and $(x-2) \to 0^-$, giving $-\infty$. As $x \to 2^+$, $\ln(x-1) \to 0^+$ and $(x-2) \to 0^+$, giving $+\infty$. Choice A is wrong because the one-sided limits don't exist. Choice B has the wrong signs. Choice D is wrong about both limits being the same.
3. The utility U(x) a consumer receives from a quantity x of a product is given by U(x) = 5 \ln(2x - 10). The function is defined only for x > 5. Determine the limiting utility as consumption x approaches the minimum required quantity of 5 units from the right.
   Answer: -\infty We need to evaluate the limit \lim_{x \to 5^+} 5 \ln(2x - 10). Let u = 2x - 10. As x approaches 5 from the right (x \to 5^+), x is slightly larger than 5, so 2x is slightly larger than 10, and u = 2x-10 approaches 0 from the positive side (u \to 0^+). The problem becomes finding \lim_{u \to 0^+} 5 \ln(u). The natural logarithm function \ln(u) approaches -\infty as its argument u approaches 0^+. Therefore, the limit is 5 \times (-\infty) = -\infty.
4. The value of a company's stock, V(t) in dollars, is modeled by the function below, where t is the number of hours from the market opening. At t=2, a major negative earnings report is released. $V(t) = \begin{cases} t^2 + 5t + 50, & \text{if } 0 \le t < 2 \\ \frac{150t - 240}{t}, & \text{if } t \ge 2 \end{cases}$ What is the instantaneous change in the stock's value due to the report?
   Answer: A decrease of $34 The instantaneous change is the difference between the value approached immediately after the report and the value approached immediately before it. This is \lim_{t \to 2^+} V(t) - \lim_{t \to 2^-} V(t). First, find the limit from the left (before the report): \lim_{t \to 2^-} V(t) = \lim_{t \to 2^-} (t^2 + 5t + 50) = (2)^2 + 5(2) + 50 = 4 + 10 + 50 = 64. Next, find the limit from the right (after the report): \lim_{t \to 2^+} V(t) = \lim_{t \to 2^+} \frac{150t - 240}{t} = \frac{150(2) - 240}{2} = \frac{300 - 240}{2} = \frac{60}{2} = 30. The change in value is $30 - $64 = -$34. This represents a decrease of $34.
5. A model for the market saturation S(t) of a product t months after a viral marketing campaign begins at t=0 is given by S(t) = \frac{1}{1 + 50e^{10/t}} for t > 0. Which statement describes the market saturation as the campaign is just launched?
   Answer: The market saturation approaches 0. To find the market saturation as the campaign is just launched, we need to evaluate the limit as t approaches 0 from the right (since time t must be positive): \lim_{t \to 0^+} S(t). First, consider the exponent 10/t. As t \to 0^+, 10/t \to +\infty. Next, consider the exponential term e^{10/t}. As the exponent approaches +\infty, e^{10/t} also approaches +\infty. Then, the denominator 1 + 50e^{10/t} approaches 1 + 50(\infty) = +\infty. Finally, the entire expression S(t) = \frac{1}{1 + 50e^{10/t}} approaches \frac{1}{+\infty}, which is 0.
6. For the piecewise function $g(x) = \begin{cases} \frac{x^2 - 9}{x + 3} & \text{if } x < -3 \\ \frac{2x + 1}{x + 3} & \text{if } x > -3 \end{cases}$, what are the left-hand and right-hand limits as $x$ approaches $-3$?
   Answer: Left-hand limit is $-6$ and right-hand limit is $-\infty$ due to sign analysis of the denominator For the left-hand limit ($x \to -3^-$): $\frac{x^2-9}{x+3} = \frac{(x-3)(x+3)}{x+3} = x-3$ for $x \neq -3$. So $\lim_{x \to -3^-} g(x) = -3-3 = -6$. For the right-hand limit ($x \to -3^+$): $\frac{2x+1}{x+3}$. As $x \to -3^+$, numerator approaches $2(-3)+1 = -5$ and denominator approaches $0^+$. Therefore $\lim_{x \to -3^+} g(x) = -\infty$. Choice A is wrong about the right-hand limit sign. Choice C is wrong because the pieces don't both simplify to removable discontinuities. Choice D has the wrong signs for both limits.
7. Let f(x) = \frac{x^2 - 4}{x^2 - (k+2)x + 2k}. The graph of y=f(x) has exactly one vertical asymptote, and k > 2. Find \lim_{x \to k^-} f(x).
   Answer: - \infty First, factor the numerator and denominator. Numerator: x^2-4 = (x-2)(x+2). Denominator: x^2 - (k+2)x + 2k = (x-k)(x-2). So, f(x) = \frac{(x-2)(x+2)}{(x-k)(x-2)}. For the function to have exactly one vertical asymptote, there must be a common factor that cancels, leaving only one factor in the denominator. The (x-2) term cancels, provided k \neq 2, which is guaranteed by the condition k>2. The simplified function is g(x) = \frac{x+2}{x-k} for x \neq 2. The single vertical asymptote is at x=k. We need to find \lim_{x \to k^-} \frac{x+2}{x-k}. As x \to k^-, x is slightly less than k. - The numerator x+2 approaches k+2. Since k>2, k+2 is positive. - The denominator x-k approaches 0 from the negative side (0^-). The limit is \frac{k+2}{0^-} = \frac{\text{positive}}{0^-} = -\infty.
8. A company finds its daily cost C(x) to produce x units changes abruptly due to a technology switch at x=50. The cost function is C(x) = \frac{x^2 + 50x - 5000}{|x - 50|}. Evaluate the limit of the cost as production approaches 50 units from above.
   Answer: $150 We need to find \lim_{x \to 50^+} C(x). First, factor the numerator: x^2 + 50x - 5000 = (x+100)(x-50). So, C(x) = \frac{(x+100)(x-50)}{|x - 50|}. As x approaches 50 from above (x \to 50^+), x is slightly greater than 50, so x-50 is a small positive number. Therefore, |x-50| = x-50. The expression becomes \lim_{x \to 50^+} \frac{(x+100)(x-50)}{x-50}. We can cancel the (x-50) terms, leaving \lim_{x \to 50^+} (x+100) = 50 + 100 = 150.
9. A manufacturing cost function is given by $C(q) = \frac{5q^2 + 20q}{q^2 - 4q}$ where $q$ is the quantity produced in thousands of units. What happens to the cost per unit as production approaches 4 thousand units?
   Answer: The cost approaches $-\infty$ from the left and $+\infty$ from the right due to sign changes in the denominator First factor: $C(q) = \frac{5q(q+4)}{q(q-4)} = \frac{5(q+4)}{q-4}$ for $q \neq 0$. At $q = 4$: numerator = $5(8) = 40 \neq 0$, denominator = $0$. For $q \to 4^-$: $(q-4) < 0$, so $C(q) \to \frac{40}{0^-} = -\infty$. For $q \to 4^+$: $(q-4) > 0$, so $C(q) \to \frac{40}{0^+} = +\infty$. Choice A is wrong because there's no factor cancellation eliminating the zero in the denominator. Choice B is wrong about the direction from each side. Choice D has the signs reversed.
10. The total cost, in dollars, to produce q custom-designed circuit boards is C(q) = 5000 + 120q + 0.1q^2. The average cost per board is given by AC(q) = C(q)/q. Which statement best describes the behavior of the average cost as the number of boards produced approaches zero?
    Answer: The average cost per board increases without bound. The average cost function is AC(q) = \frac{5000 + 120q + 0.1q^2}{q}. We can rewrite this as AC(q) = \frac{5000}{q} + 120 + 0.1q. To find the behavior as q approaches zero, we take the limit from the right (since production q must be non-negative): \lim_{q \to 0^+} AC(q) = \lim_{q \to 0^+} (\frac{5000}{q} + 120 + 0.1q). As q approaches 0 from the positive side, the term \frac{5000}{q} approaches +\infty. The other terms approach 120 and 0. Therefore, the entire expression approaches +\infty, meaning the average cost increases without bound.
11. Let f(x) = \frac{kx + 8}{x - 4}. If k is a constant such that \lim_{x \to 4^+} f(x) = -\infty, which of the following must be true?
    Answer: k < -2 We are given \lim_{x \to 4^+} \frac{kx + 8}{x - 4} = -\infty. As x approaches 4 from the right (x \to 4^+), the denominator x-4 is a small positive number, approaching 0^+. For the overall limit of the fraction to be -\infty, the numerator must approach a finite negative number. We evaluate the numerator at x=4: k(4) + 8 = 4k + 8. This value must be negative. So, we set up the inequality 4k + 8 < 0. Solving for k: 4k < -8, which simplifies to k < -2.
12. Consider the function $f(x) = \frac{x^2 - 4x + 4}{x - 2}$. If the function is modified by removing the point where $x = 2$ from its domain, what can be concluded about the one-sided limits at $x = 2$?
    Answer: Both one-sided limits exist and are equal, indicating a removable discontinuity at $x = 2$ First, factor the numerator: $x^2 - 4x + 4 = (x-2)^2$. So $f(x) = \frac{(x-2)^2}{x-2} = x-2$ for $x \neq 2$. Both $\lim_{x \to 2^-} f(x) = 0$ and $\lim_{x \to 2^+} f(x) = 0$. Since both one-sided limits exist and are equal, this indicates a removable discontinuity. Choice B is wrong because there's no sign change in the denominator after cancellation. Choice C is wrong because after simplification, there's no infinite behavior. Choice D is wrong because the simplified function doesn't oscillate.
13. The concentration C(t) of a drug in the bloodstream, in mg/L, t hours after administration is modeled by C(t) = \frac{2t - 4}{t^2 - 5t + 4} for t \ge 0. Which statement describes the behavior of the concentration as time approaches 4 hours from below?
    Answer: The concentration decreases without bound (-\infty). We need to evaluate the limit \lim_{t \to 4^-} C(t). First, factor the denominator of the function: C(t) = \frac{2t - 4}{(t-4)(t-1)}. As t approaches 4 from the left (t \to 4^-), t is slightly less than 4. - The numerator 2t - 4 approaches 2(4) - 4 = 4. - The term t-1 in the denominator approaches 4-1=3. - The term t-4 in the denominator approaches 0 from the negative side (e.g., 3.9 - 4 = -0.1), which we denote as 0^-. The limit is \frac{4}{(0^-)(3)} = \frac{4}{0^-}, which approaches -\infty. A negative concentration is not physically possible, but it is the mathematical result of this model.
14. Analyze the function f(x) = \frac{x-3}{\sqrt{x^2 - 9}}. Find \lim_{x \to -3^-} f(x).
    Answer: -\infty The domain of the function is where x^2 - 9 > 0, which means x^2 > 9, so x > 3 or x < -3. The limit as x approaches -3 from the left (x \to -3^-) is therefore valid to consider. As x \to -3^-, x is slightly less than -3 (e.g., -3.1). 1. The numerator x-3 approaches -3 - 3 = -6. 2. The term x^2 approaches (-3)^2 = 9 from the right side (since (-3.1)^2 = 9.61 > 9). We can denote this as x^2 \to 9^+. 3. The expression under the square root, x^2 - 9, approaches 9^+ - 9 = 0^+. 4. The denominator \sqrt{x^2 - 9} approaches \sqrt{0^+} which is 0^+. Combining these, the limit is \frac{-6}{0^+} which evaluates to -\infty.
15. A company's marginal profit MP(q) (in thousands of dollars per unit) for producing q units is given by the piecewise function below. What is the limiting marginal profit as production approaches 3 units from the left? $MP(q) = \begin{cases} 2q - q^2, & \text{if } 0 \le q < 3 \\ \frac{15-2q}{q-2}, & \text{if } q \ge 3 \end{cases}$
    Answer: $-3,000 To find the limit as q approaches 3 from the left (q \to 3^-), we must use the part of the function defined for q < 3, which is MP(q) = 2q - q^2. We then evaluate this expression at q=3: \lim_{q \to 3^-} MP(q) = \lim_{q \to 3^-} (2q - q^2) = 2(3) - (3)^2 = 6 - 9 = -3. Since MP(q) is in thousands of dollars, the limiting marginal profit is -$3,000.
16. The function $k(x) = \frac{e^x - 1}{x \sin(x)}$ exhibits interesting behavior near $x = 0$. Using the fact that $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$, what can be determined about the one-sided limits of $k(x)$ at $x = 0$?
    Answer: Both one-sided limits equal 1, indicating a removable discontinuity that can be filled by defining $k(0) = 1$ We can rewrite $k(x) = \frac{e^x - 1}{x \sin(x)} = \frac{e^x - 1}{x} \cdot \frac{1}{\sin(x)}$. As $x \to 0$, we have $\frac{e^x - 1}{x} \to 1$ and $\frac{1}{\sin(x)} = \frac{x}{\sin(x)} \cdot \frac{1}{x} \to 1 \cdot \frac{1}{x}$. Wait, this approach is getting complicated. Better approach: $k(x) = \frac{e^x - 1}{x} \cdot \frac{1}{\sin(x)} \cdot x = \frac{e^x - 1}{x} \cdot \frac{x}{\sin(x)}$. Using the given limits, this approaches $1 \cdot 1 = 1$ as $x \to 0$ from either side. Choice B is wrong because sine doesn't create different one-sided behavior here. Choice C is wrong about infinite limits. Choice D is wrong about non-existence.
17. Consider the piecewise-defined function $m(x) = \begin{cases} \frac{x^3 - 8}{x - 2} & \text{if } x < 2 \\ \frac{4x - 8}{x - 2} & \text{if } x > 2 \end{cases}$. What is the relationship between the left-hand and right-hand limits as $x$ approaches 2?
    Answer: The left-hand limit is 12 and the right-hand limit is 4, creating a jump discontinuity of magnitude 8 For the left-hand limit: $\frac{x^3 - 8}{x - 2} = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2 + 2x + 4$ for $x \neq 2$. So $\lim_{x \to 2^-} m(x) = 4 + 4 + 4 = 12$. For the right-hand limit: $\frac{4x - 8}{x - 2} = \frac{4(x-2)}{x-2} = 4$ for $x \neq 2$. So $\lim_{x \to 2^+} m(x) = 4$. The limits are different (12 vs 4), creating a jump discontinuity. Choice A is wrong because the right limit is 4, not 12. Choice C reverses the values. Choice D is wrong because both pieces have removable discontinuities when considered separately.

Start this quiz