Business Calculus Quiz: One Sided And Infinite Limits
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One Sided And Infinite LimitsQuestion 1 of 17

Consider the function f(x)=x2−4x+4x−2f(x) = \frac{x^2 - 4x + 4}{x - 2}. If the function is modified by removing the point where x=2x = 2 from its domain, what can be concluded about the one-sided limits at x=2x = 2?

Both one-sided limits exist and are equal, indicating a removable discontinuity at x=2x = 2
The left-hand limit approaches −∞-\infty while the right-hand limit approaches +∞+\infty
Both one-sided limits approach +∞+\infty, creating a vertical asymptote at x=2x = 2
The one-sided limits do not exist because the function oscillates near x=2x = 2
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Business Calculus Quiz

Business Calculus Quiz: One Sided And Infinite Limits

Practice One Sided And Infinite Limits in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One Sided And Infinite Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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Question 1

Consider the function f(x)=x2−4x+4x−2f(x) = \frac{x^2 - 4x + 4}{x - 2}. If the function is modified by removing the point where x=2x = 2 from its domain, what can be concluded about the one-sided limits at x=2x = 2?

  1. Both one-sided limits exist and are equal, indicating a removable discontinuity at x=2x = 2 (correct answer)
  2. The left-hand limit approaches −∞-\infty while the right-hand limit approaches +∞+\infty
  3. Both one-sided limits approach +∞+\infty, creating a vertical asymptote at x=2x = 2
  4. The one-sided limits do not exist because the function oscillates near x=2x = 2
Explanation: First, factor the numerator: x2−4x+4=(x−2)2x^2 - 4x + 4 = (x-2)^2. So f(x)=(x−2)2x−2=x−2f(x) = \frac{(x-2)^2}{x-2} = x-2 for x≠2x \neq 2. Both lim⁡x→2−f(x)=0\lim_{x \to 2^-} f(x) = 0 and lim⁡x→2+f(x)=0\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.

Question 2

A manufacturing cost function is given by C(q)=5q2+20qq2−4qC(q) = \frac{5q^2 + 20q}{q^2 - 4q} where qq is the quantity produced in thousands of units. What happens to the cost per unit as production approaches 4 thousand units?

  1. The cost approaches a finite limit of 204=5\frac{20}{4} = 5 dollars per unit due to factor cancellation in the rational expression
  2. The cost approaches +∞+\infty from both sides because both numerator and denominator approach zero simultaneously
  3. The cost approaches −∞-\infty from the left and +∞+\infty from the right due to sign changes in the denominator (correct answer)
  4. The cost approaches +∞+\infty from the left and −∞-\infty from the right due to the factored form of the expression
Explanation: First factor: C(q)=5q(q+4)q(q−4)=5(q+4)q−4C(q) = \frac{5q(q+4)}{q(q-4)} = \frac{5(q+4)}{q-4} for q≠0q \neq 0. At q=4q = 4: numerator = 5(8)=40≠05(8) = 40 \neq 0, denominator = 00. For q→4−q \to 4^-: (q−4)<0(q-4) < 0, so C(q)→400−=−∞C(q) \to \frac{40}{0^-} = -\infty. For q→4+q \to 4^+: (q−4)>0(q-4) > 0, so C(q)→400+=+∞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.

Question 3

The function k(x)=ex−1xsin⁡(x)k(x) = \frac{e^x - 1}{x \sin(x)} exhibits interesting behavior near x=0x = 0. Using the fact that lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1 and lim⁡x→0ex−1x=1\lim_{x \to 0} \frac{e^x - 1}{x} = 1, what can be determined about the one-sided limits of k(x)k(x) at x=0x = 0?

  1. Both one-sided limits equal 1, indicating a removable discontinuity that can be filled by defining k(0)=1k(0) = 1 (correct answer)
  2. The left-hand limit is −1-1 and the right-hand limit is 11 due to the oscillatory nature of sine function
  3. Both one-sided limits are infinite because the denominator contains the product xsin⁡(x)x \sin(x) which approaches zero faster than the numerator
  4. The limits do not exist because the exponential and trigonometric functions create conflicting behavior near zero
Explanation: We can rewrite k(x)=ex−1xsin⁡(x)=ex−1x⋅1sin⁡(x)k(x) = \frac{e^x - 1}{x \sin(x)} = \frac{e^x - 1}{x} \cdot \frac{1}{\sin(x)}. As x→0x \to 0, we have ex−1x→1\frac{e^x - 1}{x} \to 1 and 1sin⁡(x)=xsin⁡(x)⋅1x→1⋅1x\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)=ex−1x⋅1sin⁡(x)⋅x=ex−1x⋅xsin⁡(x)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⋅1=11 \cdot 1 = 1 as x→0x \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.

Question 4

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))?

  1. +\infty (correct answer)
  2. -1
  3. -\infty
  4. 1
Explanation: 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.

Question 5

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.

  1. $-150
  2. $150 (correct answer)
  3. $0
  4. The limit does not exist because the one-sided limits differ.
Explanation: 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.

Question 6

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?

  1. The average cost per board approaches the marginal cost of $120.
  2. The average cost per board approaches the fixed cost of $5000.
  3. The average cost per board increases without bound. (correct answer)
  4. The average cost per board approaches $0.
Explanation: The average cost function is AC(q) = \frac{5000 + 120q + 0.1q^2}{q}. We can rewrite this as AC(q) = \frac{5000}{q} + 120 + 0.1q. To find the behavior as q approaches zero, we take the limit from the right (since production q must be non-negative): \lim_{q \to 0^+} AC(q) = \lim_{q \to 0^+} ($\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.

Question 7

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.

  1. 0
  2. +\infty
  3. -\infty (correct answer)
  4. The limit does not exist because \ln(0) is undefined.
Explanation: 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.

Question 8

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.

What is the instantaneous change in the stock's value due to the report?

  1. A decrease of $34 (correct answer)
  2. An increase of $34
  3. A decrease of $30
  4. A decrease of $20
Explanation: The instantaneous change is the difference between the value approached immediately after the report and the value approached immediately before it. This is \lim_{t \to 2^+} V(t) - \lim_{t \to 2^-} V(t). First, find the limit from the left (before the report): \lim_{t \to 2^-} V(t) = \lim_{t \to 2^-} ($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.

Question 9

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?

  1. The market saturation approaches 0. (correct answer)
  2. The market saturation approaches 1.
  3. The market saturation approaches 1/51.
  4. The market saturation increases without bound.
Explanation: 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.

Question 10

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?

  1. The concentration approaches 0 mg/L.
  2. The concentration increases without bound (+$\infty$).
  3. The concentration decreases without bound (-$\infty$). (correct answer)
  4. The concentration approaches 2/3 mg/L.
Explanation: 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.

Question 11

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?

  1. $9,000
  2. $3,000
  3. The limit does not exist.
  4. $-3,000 (correct answer)
Explanation: 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.

Question 12

Consider the function h(x)=ln⁡(x−1)x−2h(x) = \frac{\ln(x-1)}{x-2}. What is the behavior of this function as xx approaches 2 from the left and from the right?

  1. Left-hand limit is 00 and right-hand limit is 00, creating a removable discontinuity at x=2x = 2
  2. Left-hand limit is +∞+\infty and right-hand limit is −∞-\infty due to the logarithmic numerator behavior
  3. Left-hand limit is −∞-\infty and right-hand limit is +∞+\infty due to denominator sign change (correct answer)
  4. Both one-sided limits approach +∞+\infty because the logarithmic function dominates near the discontinuity
Explanation: As x→2x \to 2, the numerator ln⁡(x−1)→ln⁡(1)=0\ln(x-1) \to \ln(1) = 0 and denominator (x−2)→0(x-2) \to 0. This is a 00\frac{0}{0} form requiring L'Hôpital's rule or careful analysis. Using L'Hôpital's rule: lim⁡x→2ln⁡(x−1)x−2=lim⁡x→21x−11=12−1=1\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→2−x \to 2^-, ln⁡(x−1)→0+\ln(x-1) \to 0^+ and (x−2)→0−(x-2) \to 0^-, giving −∞-\infty. As x→2+x \to 2^+, ln⁡(x−1)→0+\ln(x-1) \to 0^+ and (x−2)→0+(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.

Question 13

Consider the piecewise-defined function m(x)={x3−8x−2if x<24x−8x−2if x>2m(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 xx approaches 2?

  1. Both limits equal 12, so the function has a removable discontinuity that could be repaired by defining m(2)=12m(2) = 12
  2. The left-hand limit is 12 and the right-hand limit is 4, creating a jump discontinuity of magnitude 8 (correct answer)
  3. The left-hand limit is 4 and the right-hand limit is 12, creating a jump discontinuity with the right side higher
  4. Both pieces create infinite limits due to the common denominator (x−2)(x-2), making the function undefined at x=2x = 2
Explanation: For the left-hand limit: x3−8x−2=(x−2)(x2+2x+4)x−2=x2+2x+4\frac{x^3 - 8}{x - 2} = \frac{(x-2)(x^2+2x+4)}{x-2} = x^2 + 2x + 4 for x≠2x \neq 2. So lim⁡x→2−m(x)=4+4+4=12\lim_{x \to 2^-} m(x) = 4 + 4 + 4 = 12. For the right-hand limit: 4x−8x−2=4(x−2)x−2=4\frac{4x - 8}{x - 2} = \frac{4(x-2)}{x-2} = 4 for x≠2x \neq 2. So lim⁡x→2+m(x)=4\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.

Question 14

For the piecewise function g(x)={x2−9x+3if x<−32x+1x+3if x>−3g(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 xx approaches −3-3?

  1. Left-hand limit is −6-6 and right-hand limit is +∞+\infty due to different rational expressions
  2. Left-hand limit is −6-6 and right-hand limit is −∞-\infty due to sign analysis of the denominator (correct answer)
  3. Both limits are −6-6 since both pieces simplify to the same linear function near x=−3x = -3
  4. Left-hand limit is +∞+\infty and right-hand limit is −∞-\infty due to the discontinuous nature of piecewise functions
Explanation: For the left-hand limit (x→−3−x \to -3^-): x2−9x+3=(x−3)(x+3)x+3=x−3\frac{x^2-9}{x+3} = \frac{(x-3)(x+3)}{x+3} = x-3 for x≠−3x \neq -3. So lim⁡x→−3−g(x)=−3−3=−6\lim_{x \to -3^-} g(x) = -3-3 = -6. For the right-hand limit (x→−3+x \to -3^+): 2x+1x+3\frac{2x+1}{x+3}. As x→−3+x \to -3^+, numerator approaches 2(−3)+1=−52(-3)+1 = -5 and denominator approaches 0+0^+. Therefore lim⁡x→−3+g(x)=−∞\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.

Question 15

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?

  1. k > -2
  2. k = -2
  3. k < -2 (correct answer)
  4. k < 2
Explanation: 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.

Question 16

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).

  1. + \infty
  2. - \infty (correct answer)
  3. \frac{4}{2-k}
  4. 0
Explanation: 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.

Question 17

Analyze the function f(x) = \frac{x-3}{\sqrt{x^2 - 9}}. Find \lim_{x \to -3^-} f(x).

  1. 0
  2. +\infty
  3. Does not exist
  4. -\infty (correct answer)
Explanation: 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.