Business Calculus Quiz: Types Of Discontinuities
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Types Of DiscontinuitiesQuestion 1 of 17

The function f(x)=tan⁡(x)xf(x) = \frac{\tan(x)}{x} has a discontinuity at x=0x = 0. If we want to make this function continuous at x=0x = 0 by defining f(0)=kf(0) = k for some constant kk, what value of kk is required?

k=0k = 0 because tan⁡(0)=0\tan(0) = 0 and we can apply L'Hôpital's rule to the 00\frac{0}{0} form
No value of kk works because tangent has infinite discontinuities that cannot be removed
k=1k = 1 because lim⁡x→0tan⁡(x)x=lim⁡x→0sin⁡(x)x⋅1cos⁡(x)=1⋅1=1\lim_{x \to 0} \frac{\tan(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)} = 1 \cdot 1 = 1
k=πk = \pi because the period of tangent creates a scaling factor in the limit calculation
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Business Calculus Quiz

Business Calculus Quiz: Types Of Discontinuities

Practice Types Of Discontinuities in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Types Of Discontinuities, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The function f(x)=tan⁡(x)xf(x) = \frac{\tan(x)}{x} has a discontinuity at x=0x = 0. If we want to make this function continuous at x=0x = 0 by defining f(0)=kf(0) = k for some constant kk, what value of kk is required?

  1. k=0k = 0 because tan⁡(0)=0\tan(0) = 0 and we can apply L'Hôpital's rule to the 00\frac{0}{0} form
  2. No value of kk works because tangent has infinite discontinuities that cannot be removed
  3. k=1k = 1 because lim⁡x→0tan⁡(x)x=lim⁡x→0sin⁡(x)x⋅1cos⁡(x)=1⋅1=1\lim_{x \to 0} \frac{\tan(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)} = 1 \cdot 1 = 1 (correct answer)
  4. k=πk = \pi because the period of tangent creates a scaling factor in the limit calculation
Explanation: When you encounter a function that's undefined at a point but you need to make it continuous, you're dealing with a removable discontinuity. The key is finding the limit as x approaches that point. Here, f(x)=tan⁡(x)xf(x) = \frac{\tan(x)}{x} is undefined at x=0x = 0 because both the numerator and denominator equal zero, creating the indeterminate form 00\frac{0}{0}. To find the appropriate value of kk, we need lim⁡x→0tan⁡(x)x\lim_{x \to 0} \frac{\tan(x)}{x}. Since tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)}, we can rewrite this as: lim⁡x→0tan⁡(x)x=lim⁡x→0sin⁡(x)xcos⁡(x)=lim⁡x→0sin⁡(x)x⋅1cos⁡(x)\lim_{x \to 0} \frac{\tan(x)}{x} = \lim_{x \to 0} \frac{\sin(x)}{x \cos(x)} = \lim_{x \to 0} \frac{\sin(x)}{x} \cdot \frac{1}{\cos(x)} Using the fundamental limit lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1 and the fact that cos⁡(0)=1\cos(0) = 1, we get 1⋅1=11 \cdot 1 = 1. Therefore, k=1k = 1. Answer A incorrectly suggests using L'Hôpital's rule would give 0, but L'Hôpital's rule would actually yield lim⁡x→0sec⁡2(x)1=1\lim_{x \to 0} \frac{\sec^2(x)}{1} = 1. Answer B is wrong because while tangent has discontinuities at x=π2+nπx = \frac{\pi}{2} + n\pi, the discontinuity at x=0x = 0 is removable since the limit exists. Answer D incorrectly brings in the period of tangent, which is irrelevant to this local limit calculation. Study tip: When you see 00\frac{0}{0} forms involving trig functions at x=0x = 0, look for ways to use the standard limits lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1 and lim⁡x→0cos⁡(x)=1\lim_{x \to 0} \cos(x) = 1.

Question 2

The average profit function for a new product is given by $A(x) = \frac{x^2 - 16}{x^2 - kx + 8}$, where $x$ is the number of units sold in thousands. If the function has a removable discontinuity at $x=4$, what is the value of $k$?

  1. k=−4k = -4
  2. k=2k = 2
  3. k=6k = 6 (correct answer)
  4. k=8k = 8
Explanation: A rational function has a removable discontinuity at $x=a$ if both the numerator and the denominator are zero at $x=a$.
  1. Check the numerator at $x=4$: $4^2 - 16 = 16 - 16 = 0$. The condition is met for the numerator.
  2. For the discontinuity to be removable, the denominator must also be zero at $x=4$. Set the denominator equal to zero: $x^2 - kx + 8 = 0$.
  3. Substitute $x=4$ into the expression: $4^2 - k(4) + 8 = 0$.
  4. Solve for $k$: $16 - 4k + 8 = 0 \implies 24 - 4k = 0 \implies 4k = 24 \implies k = 6$. When $k=6$, the factor $(x-4)$ can be cancelled from the numerator and denominator.

Question 3

A company's manufacturing cost, $C(x)$, for producing $x$ items is given by the piecewise function: C(x)={15x+500if 0≤x≤10012x+kif x>100C(x) = \begin{cases} 15x + 500 & \text{if } 0 \le x \le 100 \\ 12x + k & \text{if } x > 100 \end{cases} For the function to be continuous at $x=100$, which would prevent a sudden jump in cost at that production level, what must be the value of the constant $k$?

  1. The function has a non-removable jump discontinuity regardless of the value of $k$.
  2. The value of $k$ must be 300300.
  3. The value of $k$ must be 500500.
  4. The value of $k$ must be 800800. (correct answer)
Explanation: For the cost function $C(x)$ to be continuous at $x=100$, the left-hand limit must equal the right-hand limit.
  1. Calculate the left-hand limit: $\lim_{x \to 100^-} C(x) = \lim_{x \to 100^-} (15x + 500) = 15(100) + 500 = 1500 + 500 = 2000$.
  2. Calculate the right-hand limit: $\lim_{x \to 100^+} C(x) = \lim_{x \to 100^+} (12x + k) = 12(100) + k = 1200 + k$.
  3. Set the limits equal to each other: $2000 = 1200 + k$.
  4. Solve for $k$: $k = 2000 - 1200 = 800$. Thus, $k$ must be 800800 for the function to be continuous.

Question 4

A function modeling a company's market share percentage over time $t$ (in years) is given by $f(t) = \frac{t^2-t-12}{t^2-9}$. Which statement accurately describes the discontinuities of this function for $t>0$?

  1. An infinite discontinuity at $t=3$ and a removable discontinuity at $t=4$.
  2. A removable discontinuity at $t=3$ and an infinite discontinuity at $t=-3$.
  3. An infinite discontinuity at $t=3$ and no other discontinuities for $t>0$. (correct answer)
  4. A removable discontinuity at $t=4$ and an infinite discontinuity at $t=3$.
Explanation: First, factor the numerator and the denominator of the function $f(t) = \frac{t^2-t-12}{t^2-9}$. Numerator: $t^2-t-12 = (t-4)(t+3)$. Denominator: $t^2-9 = (t-3)(t+3)$. So, $f(t) = \frac{(t-4)(t+3)}{(t-3)(t+3)}$. Potential discontinuities exist where the denominator is zero: $t=3$ and $t=-3$. Since the question specifies $t>0$, we only need to consider $t=3$. At $t=-3$, the factor $(t+3)$ would cancel, creating a removable discontinuity, but this is outside the domain of interest. At $t=3$, the denominator is zero, but the numerator is $(3-4)(3+3) = (-1)(6) = -6$. Since the limit approaches $\frac{-6}{0}$, this is an infinite discontinuity (a vertical asymptote). The point $t=4$ makes the numerator zero, but this is a root of the function, not a discontinuity. Therefore, the only discontinuity for $t>0$ is an infinite one at $t=3$.

Question 5

The cost, in dollars, to remove $p$ percent of a pollutant from a factory's emissions is given by the function $C(p) = \frac{75000p}{100-p}$ for $0 \le p < 100$. Analyze the behavior of this cost function as $p$ approaches 100100.

  1. The function has a removable discontinuity at $p=100$, meaning the cost for 100% removal can be defined by a special contract.
  2. The function has an infinite discontinuity at $p=100$, implying the cost becomes prohibitively large as removal approaches 100%. (correct answer)
  3. The function has a jump discontinuity at $p=100$, indicating a sudden fixed charge for achieving 100% pollutant removal.
  4. The function is continuous at $p=100$, meaning the cost of 100% removal is a finite, calculable amount.
Explanation: We need to analyze the limit of the cost function $C(p)$ as $p$ approaches 100100 from the left side (since $p<100$). $\lim_{p \to 100^-} C(p) = \lim_{p \to 100^-} \frac{75000p}{100-p}$. As $p$ approaches 100100, the numerator approaches $75000(100) = 7,500,000$. The denominator $(100-p)$ approaches 00 through positive values. A non-zero number divided by a quantity approaching zero results in a limit of $\infty$ or $-\infty$. In this case, since both numerator and denominator are positive, the limit is $\infty$. This signifies an infinite discontinuity, which in this business context means the cost to remove every last particle of the pollutant would be infinitely large.

Question 6

A tiered pricing model for a software subscription is based on the number of users, $n$. The annual cost per user, $P(n)$, is given by $P(n) = 500 - 2\lfloor n/100 \rfloor$, where $\lfloor \cdot \rfloor$ is the floor function. What type of discontinuity does the total cost function, $C(n) = n \cdot P(n)$, have at $n=100$?

  1. A removable discontinuity because a special price can be set for exactly 100 users.
  2. The function is continuous because the price drop is designed to be seamless.
  3. An infinite discontinuity because the cost function is undefined for some values.
  4. A jump discontinuity because the total cost abruptly changes at 100 users. (correct answer)
Explanation: Let's analyze the total cost function $C(n) = n(500 - 2\lfloor n/100 \rfloor)$ around $n=100$. We need to check the one-sided limits.
  1. Left-hand limit: As $n \to 100^-$ (e.g., $n=99.99$), $n/100$ is slightly less than 1, so $\lfloor n/100 \rfloor = 0$. $\lim_{n \to 100^-} C(n) = \lim_{n \to 100^-} n(500 - 2(0)) = 100(500) = 50000$.
  2. Right-hand limit: As $n \to 100^+$ (e.g., $n=100.01$), $n/100$ is slightly more than 1, so $\lfloor n/100 \rfloor = 1$. The value at $n=100$ also uses $\lfloor 100/100 \rfloor = 1$. $\lim_{n \to 100^+} C(n) = \lim_{n \to 100^+} n(500 - 2(1)) = 100(498) = 49800$. Since the left-hand limit ($50,000$) and the right-hand limit ($49,800$) are different, the function has a jump discontinuity at $n=100$. The total cost for 99 users is nearly $50,000, but the cost for 100 users is $49,800.

Question 7

The profit function $P(x)$ for a certain product is known to be continuous for all $x \ge 0$, where $x$ is the number of units produced. What is the most significant business implication of this continuity?

  1. The profit is guaranteed to be positive for any number of units produced in the feasible range.
  2. The company can make small adjustments to production levels without causing sudden, drastic shocks to profit. (correct answer)
  3. There is a single, unique production level that will maximize the profit for the company.
  4. The cost to produce each additional unit remains constant across all production levels.
Explanation: Continuity of a function, in a practical sense, means that small changes in the input result in small changes in the output. For a profit function $P(x)$, this means that if the production level $x$ is changed by a small amount, the profit $P(x)$ will also change by a small amount. This implies predictability and stability; there are no hidden 'jumps' or 'holes' where producing one more unit leads to a surprisingly different profit. Choice A is incorrect; continuity doesn't guarantee positivity. Choice C refers to the existence of a maximum, which is not guaranteed by continuity alone (differentiability is usually assumed). Choice D is incorrect; continuity does not imply a constant marginal cost.

Question 8

An internet service provider charges a monthly fee based on data usage, $x$, in gigabytes. The cost function is $C(x) = \begin{cases} 30 & \text{if } 0 \le x \le 50 \\ 30 + 0.50(x-50) & \text{if } x > 50 \end{cases}$. Which statement best describes the continuity of this cost function at the $x=50$ GB threshold?

  1. It is continuous because the cost for using slightly more than 50 GB smoothly continues from the cost of using 50 GB. (correct answer)
  2. It has a removable discontinuity because the company could offer a discount at exactly 50 GB.
  3. It has a jump discontinuity because the rate changes from $0 per additional GB to $0.50 per GB.
  4. It has an infinite discontinuity because heavy users face unbounded costs.
Explanation: When analyzing piecewise functions, you need to check continuity by examining whether the function value approaches the same limit from both sides at the boundary point. To test continuity at x=50x = 50, evaluate three things: the left-hand limit, right-hand limit, and the actual function value at x=50x = 50. From the left (x→50−x \to 50^-): Using the first piece, C(50)=30C(50) = 30 From the right (x→50+x \to 50^+): Using the second piece, C(x)=30+0.50(x−50)C(x) = 30 + 0.50(x-50). As xx approaches 50, this becomes 30+0.50(50−50)=30+0=3030 + 0.50(50-50) = 30 + 0 = 30 At the point: C(50)=30C(50) = 30 (using the first piece since 50≤5050 \leq 50) Since all three values equal 30, the function is continuous at x=50x = 50. Choice A correctly identifies this continuity—the cost transitions smoothly without any sudden jumps. Choice B is wrong because there's no removable discontinuity here; the function is already continuous. Choice C misunderstands the concept—while the rate of change (derivative) changes from 0 to 0.50, this creates a corner, not a jump discontinuity. The function value itself doesn't jump. Choice D is incorrect because the function grows linearly for x>50x > 50, so costs increase predictably, not unboundedly. Study tip: Don't confuse discontinuity in the function with discontinuity in its derivative. A piecewise function can be continuous even when its rate of change suddenly shifts—focus on whether the function values themselves connect smoothly at boundary points.

Question 9

A function $h(x)$ has a discontinuity at $x=c$. It is known that $\lim_{x \to c} h(x)$ exists and is a finite value, but $h(c)$ is undefined. What conclusion can be drawn about the discontinuity at $x=c$?

  1. It is a jump discontinuity because the function is undefined at the point.
  2. It is an infinite discontinuity because there must be a vertical asymptote.
  3. It is a removable discontinuity that can be eliminated by defining $h(c)$ to equal the limit. (correct answer)
  4. It is an oscillating discontinuity because the function approaches a value but never reaches it.
Explanation: The definition of a removable discontinuity at a point $x=c$ is that the limit, $\lim_{x \to c} h(x)$, exists and is finite, but either $h(c)$ is not defined or $h(c) \neq \lim_{x \to c} h(x)$. The question states that the limit exists but $h(c)$ is undefined. This fits the definition of a removable discontinuity perfectly. It is 'removable' because one could define $h(c)$ to be equal to the limit, making the function continuous at that point.

Question 10

A company's profit function is P(t)={2t+1if t<310if t=3t2−2t+4if t>3P(t) = \begin{cases} 2t + 1 & \text{if } t < 3 \\ 10 & \text{if } t = 3 \\ t^2 - 2t + 4 & \text{if } t > 3 \end{cases} where tt represents time in years. To properly classify the discontinuity at t=3t = 3, which analysis is required?

  1. Compare lim⁡t→3−P(t)\lim_{t \to 3^-} P(t) and lim⁡t→3+P(t)\lim_{t \to 3^+} P(t) only, since the function value is irrelevant for classification
  2. Evaluate lim⁡t→3−P(t)=7\lim_{t \to 3^-} P(t) = 7, lim⁡t→3+P(t)=7\lim_{t \to 3^+} P(t) = 7, and P(3)=10P(3) = 10 to identify a removable discontinuity (correct answer)
  3. Check if the denominator approaches zero, since piecewise functions always have infinite discontinuities at boundaries
  4. Determine that no discontinuity exists because each piece is continuous on its respective domain interval
Explanation: To classify a discontinuity, we need all three values: lim⁡t→3−P(t)=2(3)+1=7\lim_{t \to 3^-} P(t) = 2(3) + 1 = 7, lim⁡t→3+P(t)=32−2(3)+4=7\lim_{t \to 3^+} P(t) = 3^2 - 2(3) + 4 = 7, and P(3)=10P(3) = 10. Since both one-sided limits exist and are equal, but the function value differs, this is a removable discontinuity. Choice A is wrong because the function value matters for classification. Choice C is wrong because piecewise functions don't automatically have infinite discontinuities. Choice D is wrong because continuity of individual pieces doesn't guarantee continuity at boundary points.

Question 11

A company's marginal cost function has the form MC(x)=x2−16(x−4)2MC(x) = \frac{x^2 - 16}{(x-4)^2}. Management wants to know if the discontinuity at x=4x = 4 can be 'repaired' by redefining the function at that point. What should an analyst conclude?

  1. The discontinuity is removable because factoring eliminates one power of (x−4)(x-4) from numerator and denominator
  2. The discontinuity is removable because limits of rational functions always exist when numerator approaches zero
  3. The discontinuity is non-removable because after factoring, (x−4)(x-4) remains in the denominator with power 1 (correct answer)
  4. The discontinuity is non-removable because the denominator has higher degree than the canceled factor in numerator
Explanation: When analyzing discontinuities in rational functions, you need to determine whether the problematic point creates a "hole" that can be filled (removable) or an asymptote that cannot be fixed (non-removable). The key is examining what happens after factoring. Let's factor the numerator: x2−16=(x+4)(x−4)x^2 - 16 = (x+4)(x-4). So our function becomes: MC(x)=(x+4)(x−4)(x−4)2=x+4x−4MC(x) = \frac{(x+4)(x-4)}{(x-4)^2} = \frac{x+4}{x-4} (for x≠4x \neq 4) After canceling one factor of (x−4)(x-4), we still have (x−4)(x-4) in the denominator with power 1. This means as xx approaches 4, the denominator approaches zero while the numerator approaches 4+4=84+4=8. This creates a vertical asymptote, making the discontinuity non-removable. Answer A incorrectly suggests that any factoring and canceling makes a discontinuity removable. While we do eliminate one power of (x−4)(x-4), what remains in the denominator determines removability. Answer B makes a false generalization about rational functions. Limits don't automatically exist just because numerators approach zero—what matters is the behavior of both numerator and denominator. Answer D focuses on degree comparison, which isn't the relevant criterion here. The key isn't comparing degrees but whether any factors remain in the denominator after canceling. Study tip: For rational function discontinuities, always factor completely and cancel common terms. If nothing remains in the denominator at the problem point, it's removable. If factors remain, it's non-removable with a vertical asymptote.

Question 12

Consider the function f(x)={x2−4x−2if x≠25if x=2f(x) = \begin{cases} \frac{x^2-4}{x-2} & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases} . A business analyst claims this function has a jump discontinuity at x=2x = 2. What is the most accurate assessment of this claim?

  1. The claim is incorrect because the function has a removable discontinuity, since the limit exists but differs from the function value (correct answer)
  2. The claim is correct because the left and right limits exist but are unequal at x=2x = 2
  3. The claim is incorrect because the function has an infinite discontinuity due to the denominator approaching zero
  4. The claim is correct because the function value at x=2x = 2 creates a vertical asymptote in the graph
Explanation: First, we find the limit: lim⁡x→2x2−4x−2=lim⁡x→2(x−2)(x+2)x−2=lim⁡x→2(x+2)=4\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4. Since lim⁡x→2f(x)=4\lim_{x \to 2} f(x) = 4 exists but f(2)=5≠4f(2) = 5 \neq 4, this is a removable discontinuity, not a jump discontinuity. Choice B is wrong because both one-sided limits equal 4. Choice C is wrong because there's no infinite discontinuity. Choice D is wrong because there's no vertical asymptote.

Question 13

The function $f(x) = \frac{x^2 - 1}{|x-1|}$ represents the efficiency rating of a production process. What type of discontinuity does $f(x)$ have at $x=1$?

  1. A removable discontinuity because $\lim_{x \to 1} f(x)$ exists.
  2. A jump discontinuity because the left-hand and right-hand limits are finite but not equal. (correct answer)
  3. An infinite discontinuity because the function's value grows without bound.
  4. The function is continuous at $x=1$ because the numerator is zero.
Explanation: To determine the type of discontinuity at $x=1$, we must evaluate the one-sided limits. The absolute value function $|x-1|$ is defined differently for $x>1$ and $x<1$.
  1. Right-hand limit ($x \to 1^+$): For $x>1$, $|x-1| = x-1$. So, $\lim_{x \to 1^+} \frac{x^2 - 1}{x-1} = \lim_{x \to 1^+} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1^+} (x+1) = 2$.
  2. Left-hand limit ($x \to 1^-$): For $x<1$, $|x-1| = -(x-1)$. So, $\lim_{x \to 1^-} \frac{x^2 - 1}{-(x-1)} = \lim_{x \to 1^-} \frac{(x-1)(x+1)}{-(x-1)} = \lim_{x \to 1^-} -(x+1) = -2$. Since the left-hand limit ($-2$) and the right-hand limit ($2$) both exist but are not equal, the function has a jump discontinuity at $x=1$.

Question 14

A company models its daily revenue, $R(t)$, as a function of time $t$ in days with $R(t) = \frac{|t-5|(t+2)}{t^2-3t-10}$. Analyze the discontinuities of this function.

  1. A jump discontinuity at $t=5$ and a removable discontinuity at $t=-2$. (correct answer)
  2. A removable discontinuity at $t=5$ and a jump discontinuity at $t=-2$.
  3. Removable discontinuities at both $t=5$ and $t=-2$.
  4. Jump discontinuities at both $t=5$ and $t=-2$.
Explanation: First, factor the denominator: $t^2-3t-10 = (t-5)(t+2)$. The function is $R(t) = \frac{|t-5|(t+2)}{(t-5)(t+2)}$. Potential discontinuities are at $t=5$ and $t=-2$.
  1. Analyze at $t=-2$: The factor $(t+2)$ appears in both the numerator and denominator. $\lim_{t \to -2} \frac{|t-5|}{t-5} = \frac{|-7|}{-7} = \frac{7}{-7} = -1$. Since the limit exists, the discontinuity at $t=-2$ is removable.
  2. Analyze at $t=5$: We need to check one-sided limits due to the absolute value $|t-5|$.
    • Right-hand limit: $\lim_{t \to 5^+} \frac{(t-5)(t+2)}{(t-5)(t+2)} = \lim_{t \to 5^+} 1 = 1$.
    • Left-hand limit: $\lim_{t \to 5^-} \frac{-(t-5)(t+2)}{(t-5)(t+2)} = \lim_{t \to 5^-} -1 = -1$. Since the left and right-hand limits exist but are not equal ($-1 \neq 1$), the discontinuity at $t=5$ is a jump discontinuity.

Question 15

A cost function C(x)=x2−9x+3+2x−1C(x) = \frac{x^2 - 9}{x + 3} + \frac{2}{x - 1} has discontinuities at multiple points. Which statement correctly characterizes these discontinuities?

  1. Both x=−3x = -3 and x=1x = 1 have infinite discontinuities because the function approaches infinity at both points
  2. The point x=1x = 1 has a removable discontinuity while x=−3x = -3 has an infinite discontinuity
  3. Both points have removable discontinuities because limits can be evaluated using algebraic manipulation
  4. The point x=−3x = -3 has a removable discontinuity while x=1x = 1 has an infinite discontinuity (correct answer)
Explanation: When analyzing discontinuities in rational functions, you need to examine what happens at points where the function is undefined and determine whether the discontinuity can be "fixed" through algebraic simplification. Let's examine each discontinuity in C(x)=x2−9x+3+2x−1C(x) = \frac{x^2 - 9}{x + 3} + \frac{2}{x - 1}: For x=−3x = -3: The first fraction becomes 00\frac{0}{0}, an indeterminate form. Factor the numerator: x2−9=(x−3)(x+3)x^2 - 9 = (x-3)(x+3). This gives us (x−3)(x+3)x+3=x−3\frac{(x-3)(x+3)}{x+3} = x-3 after canceling the common factor. As x→−3x \to -3, this approaches −3−3=−6-3-3 = -6. The second fraction 2x−1\frac{2}{x-1} approaches 2−4=−12\frac{2}{-4} = -\frac{1}{2}. Since both parts have finite limits, this is a removable discontinuity. For x=1x = 1: The first fraction x2−9x+3\frac{x^2-9}{x+3} approaches −84=−2\frac{-8}{4} = -2. However, the second fraction 2x−1\frac{2}{x-1} approaches ±∞\pm\infty (the sign depends on the direction of approach). This makes the entire function approach infinity, creating an infinite discontinuity. Answer D correctly identifies this pattern. Answer A wrongly claims both are infinite discontinuities. Answer B reverses the types of discontinuities. Answer C incorrectly suggests both are removable, missing that x=1x = 1 creates an unbounded behavior. Remember: A discontinuity is removable only if the limit exists and is finite at that point. If any part of the function approaches infinity, you have an infinite discontinuity.

Question 16

A revenue function R(x)=∣x−5∣x−5+2R(x) = \frac{|x - 5|}{x - 5} + 2 is undefined at x=5x = 5. What happens if we define R(5)=3R(5) = 3 to make the function continuous?

  1. The function becomes continuous because we've assigned a finite value that matches the two-sided limit
  2. The function remains discontinuous because the original undefined point cannot be removed by assignment
  3. The function becomes continuous because any finite value assigned at x=5x = 5 removes the discontinuity
  4. The function remains discontinuous because the left and right limits are different, creating a jump discontinuity (correct answer)
Explanation: When analyzing continuity at a point where a function is undefined, you need to examine what happens as you approach that point from both sides and determine if assigning a value can create continuity. Let's examine the behavior of R(x)=∣x−5∣x−5+2R(x) = \frac{|x - 5|}{x - 5} + 2 near x=5x = 5. The absolute value expression ∣x−5∣|x - 5| behaves differently on each side of x=5x = 5. When x<5x < 5, we have x−5<0x - 5 < 0, so ∣x−5∣=−(x−5)|x - 5| = -(x - 5). Therefore, ∣x−5∣x−5=−(x−5)x−5=−1\frac{|x - 5|}{x - 5} = \frac{-(x - 5)}{x - 5} = -1. When x>5x > 5, we have x−5>0x - 5 > 0, so ∣x−5∣=x−5|x - 5| = x - 5, giving us ∣x−5∣x−5=x−5x−5=1\frac{|x - 5|}{x - 5} = \frac{x - 5}{x - 5} = 1. This means the left-hand limit is lim⁡x→5−R(x)=−1+2=1\lim_{x \to 5^-} R(x) = -1 + 2 = 1, while the right-hand limit is lim⁡x→5+R(x)=1+2=3\lim_{x \to 5^+} R(x) = 1 + 2 = 3. Since these one-sided limits are different, no two-sided limit exists. Answer A is wrong because there is no two-sided limit to match. Answer B incorrectly suggests that undefined points can never be made continuous by assignment—they can when removable discontinuities exist. Answer C is wrong because you cannot arbitrarily assign any value to create continuity; the assigned value must equal the two-sided limit. Answer D correctly identifies that the different left and right limits create a jump discontinuity that cannot be eliminated. Remember: A function can only be made continuous at a point by assignment if the two-sided limit exists there.

Question 17

For the function $f(x) = \frac{x^2+ax-6}{x-2}$ to have a removable discontinuity at $x=2$, what must be the value of the limit $\lim_{x \to 2} f(x)$?

  1. The limit must be 55. (correct answer)
  2. The limit must be 33.
  3. The limit must be 11.
  4. The limit must be −1-1.
Explanation: When you encounter a rational function with a potential discontinuity, you're dealing with limits and continuity concepts. A removable discontinuity occurs when a function has a "hole" at a point that could theoretically be "filled in" to make the function continuous there. For f(x)=x2+ax−6x−2f(x) = \frac{x^2+ax-6}{x-2} to have a removable discontinuity at x=2x=2, the numerator must also equal zero when x=2x=2. This creates the indeterminate form 00\frac{0}{0}, which can potentially be resolved through factoring. Setting the numerator equal to zero at x=2x=2: 22+a(2)−6=02^2 + a(2) - 6 = 0, which gives us 4+2a−6=04 + 2a - 6 = 0, so a=1a = 1. With a=1a = 1, our function becomes f(x)=x2+x−6x−2f(x) = \frac{x^2+x-6}{x-2}. Factoring the numerator: x2+x−6=(x+3)(x−2)x^2+x-6 = (x+3)(x-2). Now we can simplify: f(x)=(x+3)(x−2)x−2=x+3f(x) = \frac{(x+3)(x-2)}{x-2} = x+3 (for x≠2x \neq 2). Therefore, lim⁡x→2f(x)=2+3=5\lim_{x \to 2} f(x) = 2 + 3 = 5. Looking at the wrong answers: B) 3 might result from incorrectly evaluating just the x+3x+3 part at x=0x=0. C) 1 could come from mishandling the algebra or using the wrong value of aa. D) -1 might arise from sign errors in factoring or substitution. Study tip: For removable discontinuities in rational functions, always check if both numerator and denominator are zero at the problem point. If so, factor and cancel the common terms to find the limit.