Business Calculus Quiz: Computing Limits
15 questions · exam conditions
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Computing LimitsQuestion 1 of 15

For a supply function S(p)=p2−6p+9p−3S(p) = \frac{p^2 - 6p + 9}{p - 3} where pp is the price, what can be concluded about lim⁡p→3S(p)\lim_{p \to 3} S(p)?

The limit equals 00 because the numerator approaches 00
The limit equals 33 because that's the point of approach
The limit does not exist because we get 00\frac{0}{0}
The limit equals 00 after canceling common factors
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Business Calculus Quiz

Business Calculus Quiz: Computing Limits

Practice Computing 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 Computing Limits, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

How to use this quiz

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.

All questions

Question 1

For a supply function S(p)=p2−6p+9p−3S(p) = \frac{p^2 - 6p + 9}{p - 3} where pp is the price, what can be concluded about lim⁡p→3S(p)\lim_{p \to 3} S(p)?

  1. The limit equals 00 because the numerator approaches 00
  2. The limit equals 33 because that's the point of approach
  3. The limit does not exist because we get 00\frac{0}{0}
  4. The limit equals 00 after canceling common factors (correct answer)
Explanation: Factor the numerator: p2−6p+9=(p−3)2p^2 - 6p + 9 = (p-3)^2. So S(p)=(p−3)2p−3=p−3S(p) = \frac{(p-3)^2}{p-3} = p-3 for p≠3p \neq 3. Therefore, lim⁡p→3S(p)=lim⁡p→3(p−3)=0\lim_{p \to 3} S(p) = \lim_{p \to 3} (p-3) = 0. Choice A gives the right answer but wrong reasoning (you can't conclude the limit is zero just because the numerator approaches zero). Choice B incorrectly assumes the limit equals the point of approach. Choice C fails to recognize that 00\frac{0}{0} forms can have finite limits after algebraic manipulation.

Question 2

For what value of the constant cc is the following function continuous at x=9x=9? $$ f(x) = \begin{cases} \frac{x-9}{\sqrt{x}-3} & \text{if } x \neq 9 \ c & \text{if } x = 9 \end{cases}

  1. 99
  2. 33
  3. 66 (correct answer)
  4. 16\frac{1}{6}
Explanation: When you encounter a piecewise function and need to find a constant that makes it continuous at a specific point, you're looking for the value that makes the function "connect" smoothly without any jumps or breaks. For continuity at x=9x = 9, the limit as xx approaches 9 must equal the function value at x=9x = 9. Since f(9)=cf(9) = c, you need to find lim⁡x→9x−9x−3\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3}. This limit initially gives the indeterminate form 00\frac{0}{0}, so you need to simplify. The key insight is to rationalize by multiplying by the conjugate: x−9x−3⋅x+3x+3=(x−9)(x+3)(x)2−9=(x−9)(x+3)x−9\frac{x-9}{\sqrt{x}-3} \cdot \frac{\sqrt{x}+3}{\sqrt{x}+3} = \frac{(x-9)(\sqrt{x}+3)}{(\sqrt{x})^2-9} = \frac{(x-9)(\sqrt{x}+3)}{x-9} For x≠9x \neq 9, you can cancel the (x−9)(x-9) terms, leaving x+3\sqrt{x}+3. Therefore: lim⁡x→9x−9x−3=lim⁡x→9(x+3)=9+3=6\lim_{x \to 9} \frac{x-9}{\sqrt{x}-3} = \lim_{x \to 9} (\sqrt{x}+3) = \sqrt{9}+3 = 6 So c=6c = 6. Looking at the wrong answers: A) 99 might come from just substituting x=9x = 9 without proper limit evaluation. B) 33 could result from only considering 9=3\sqrt{9} = 3 and missing the addition. D) 16\frac{1}{6} might come from incorrectly taking the reciprocal of the correct answer. Strategy tip: For piecewise continuity problems, always evaluate the limit from the "complicated" piece and set it equal to the constant. Remember to simplify indeterminate forms before substituting the point value.

Question 3

A company finds that the cost, in dollars, to remove pp percent (0≤p<1000 \le p < 100) of a certain pollutant from a river is given by the function C(p)=2000p100−pC(p) = \frac{2000p}{100 - p}

Which statement best describes the meaning of the limit lim⁡p→100−C(p)\lim_{p \to 100^-} C(p) in this context?

  1. The total cost to remove the pollutant will eventually decrease after a certain percentage has been removed.
  2. The cost to remove the last fraction of a percent of the pollutant is disproportionately high, approaching an infinite amount. (correct answer)
  3. The maximum possible cost to clean the river is $200,000.
  4. For every 1% of pollutant removed, the cost increases by a constant amount.
Explanation: When you encounter a limit approaching a vertical asymptote in a business context, you're typically analyzing what happens to costs, revenue, or other quantities as they approach some boundary or constraint. Let's evaluate lim⁡p→100−C(p)=lim⁡p→100−2000p100−p\lim_{p \to 100^-} C(p) = \lim_{p \to 100^-} \frac{2000p}{100 - p}. As pp approaches 100 from the left, the numerator 2000p2000p approaches 2000(100)=200,0002000(100) = 200,000, while the denominator (100−p)(100 - p) approaches 0 from the positive side. This creates the form 200,0000+\frac{200,000}{0^+}, which means the limit is +∞+\infty. The cost grows without bound as you try to remove closer to 100% of the pollutant. Option B correctly captures this: removing those final traces of pollutant becomes prohibitively expensive, with costs approaching infinity. This reflects real-world pollution cleanup, where removing the last 1% is often more expensive than removing the first 90%. Option A is wrong because the function is always increasing—costs never decrease. Option C misinterprets the calculation; while the numerator approaches 200,000, the overall limit is infinite, not capped at $200,000. Option D is incorrect because this isn't a linear relationship—each additional percent removed costs progressively more, not a constant amount. Study tip: When you see rational functions with restricted domains in business problems, immediately check what happens at the boundary values. Vertical asymptotes often represent real constraints where costs become infinite or processes become impossible.

Question 4

Consider the piecewise function f(x)={kx2−3if x<22x+kif x≥2f(x) = \begin{cases} kx^2 - 3 & \text{if } x < 2 \\ 2x + k & \text{if } x \ge 2 \end{cases} The value of the constant kk is chosen such that the limit lim⁡x→2f(x)\lim_{x \to 2} f(x) exists. What is the value of this limit?

  1. 193\frac{19}{3} (correct answer)
  2. 73\frac{7}{3}
  3. 11
  4. The limit cannot be determined.
Explanation: For the limit to exist at x=2x=2, the left-hand limit must equal the right-hand limit. Left-hand limit: lim⁡x→2−f(x)=lim⁡x→2−(kx2−3)=k(2)2−3=4k−3\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2 - 3) = k(2)^2 - 3 = 4k - 3. Right-hand limit: lim⁡x→2+f(x)=lim⁡x→2+(2x+k)=2(2)+k=4+k\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x + k) = 2(2) + k = 4 + k. Set them equal to find kk: 4k−3=4+k⇒3k=7⇒k=734k - 3 = 4 + k \Rightarrow 3k = 7 \Rightarrow k = \frac{7}{3}. The question asks for the value of the limit, not the value of kk. Substitute k=73k = \frac{7}{3} into either expression for the limit. Using the right-hand limit expression: 4+k=4+73=123+73=1934 + k = 4 + \frac{7}{3} = \frac{12}{3} + \frac{7}{3} = \frac{19}{3}.

Question 5

The average cost, in dollars per unit, for a manufacturing process is given by the function AC(q)=16q4+8000q2+10002q2+50\text{AC}(q) = \frac{\sqrt{16q^4 + 8000q^2} + 1000}{2q^2 + 50} where qq is the number of units produced. What is the long-run average cost as the number of units produced becomes very large?

  1. ∞\infty
  2. 88
  3. 44
  4. 22 (correct answer)
Explanation: When you encounter a question about long-run behavior in business calculus, you're looking at limits as the variable approaches infinity. This tests your understanding of how rational functions behave for very large values. To find the long-run average cost, you need to evaluate lim⁡q→∞16q4+8000q2+10002q2+50\lim_{q \to \infty} \frac{\sqrt{16q^4 + 8000q^2} + 1000}{2q^2 + 50}. The key insight is to factor out the highest powers from both the numerator and denominator. In the square root, factor out q2q^2: 16q4+8000q2=q216+8000q2\sqrt{16q^4 + 8000q^2} = q^2\sqrt{16 + \frac{8000}{q^2}}. As q→∞q \to \infty, the term 8000q2→0\frac{8000}{q^2} \to 0, so this becomes q216=4q2q^2\sqrt{16} = 4q^2. The numerator approaches 4q2+10004q^2 + 1000, and since the 10001000 becomes negligible compared to 4q24q^2, it's effectively 4q24q^2. The denominator 2q2+502q^2 + 50 is effectively 2q22q^2 for large qq. Therefore: lim⁡q→∞4q22q2=2\lim_{q \to \infty} \frac{4q^2}{2q^2} = 2 Choice A (∞\infty) would occur if the numerator's degree exceeded the denominator's degree, but they're equal. Choice B (88) mistakes the coefficient under the square root for the final answer. Choice C (44) incorrectly assumes the limit is just the coefficient of q2q^2 in the numerator without considering the denominator. Study tip: For rational function limits at infinity, compare the highest degree terms after simplifying any radicals by factoring out the highest powers.

Question 6

Evaluate the limit: lim⁡x→−∞3x−24x2+x−1\lim_{x \to -\infty} \frac{3x - 2}{\sqrt{4x^2 + x - 1}}

  1. −32-\frac{3}{2} (correct answer)
  2. 32\frac{3}{2}
  3. 34\frac{3}{4}
  4. −∞-\infty
Explanation: To evaluate this limit at negative infinity, we divide the numerator and the denominator by the highest power of xx in the denominator. The term with the highest power in the denominator is 4x2\sqrt{4x^2}, which behaves like 4x2=2∣x∣\sqrt{4}\sqrt{x^2} = 2|x|. Since x→−∞x \to -\infty, xx is negative, so ∣x∣=−x|x| = -x. We divide the numerator and denominator by xx. In the denominator, we use the fact that for x<0x < 0, x=−x2x = -\sqrt{x^2}. lim⁡x→−∞3x−2x4x2+x−1x=lim⁡x→−∞3−2x−4x2+x−1x2=lim⁡x→−∞3−2x−4+1x−1x2\lim_{x \to -\infty} \frac{\frac{3x-2}{x}}{\frac{\sqrt{4x^2+x-1}}{x}} = \lim_{x \to -\infty} \frac{3 - \frac{2}{x}}{-\sqrt{\frac{4x^2+x-1}{x^2}}} = \lim_{x \to -\infty} \frac{3 - \frac{2}{x}}{-\sqrt{4 + \frac{1}{x} - \frac{1}{x^2}}} As x→−∞x \to -\infty, the terms with xx in the denominator approach 0. 3−0−4+0−0=3−2=−32\frac{3 - 0}{-\sqrt{4 + 0 - 0}} = \frac{3}{-2} = -\frac{3}{2}

Question 7

Evaluate the following limit: lim⁡x→3x−31x−13\lim_{x \to 3} \frac{x-3}{\frac{1}{x} - \frac{1}{3}}

  1. −9-9 (correct answer)
  2. 99
  3. 00
  4. The limit does not exist.
Explanation: The limit results in the indeterminate form 00\frac{0}{0}. To evaluate it, we simplify the complex fraction in the denominator: 1x−13=3−x3x\frac{1}{x} - \frac{1}{3} = \frac{3-x}{3x} The expression becomes: lim⁡x→3x−33−x3x=lim⁡x→3(x−3)⋅3x3−x\lim_{x \to 3} \frac{x-3}{\frac{3-x}{3x}} = \lim_{x \to 3} (x-3) \cdot \frac{3x}{3-x} We can factor a −1-1 from (3−x)(3-x) to get −(x−3)-(x-3): lim⁡x→3(x−3)⋅3x−(x−3)=lim⁡x→3−3x\lim_{x \to 3} \frac{(x-3) \cdot 3x}{-(x-3)} = \lim_{x \to 3} -3x Now, by direct substitution: −3(3)=−9-3(3) = -9.

Question 8

Evaluate the following limit, which represents the instantaneous rate of change of a function at a point: lim⁡h→01(3+h)2−19h\lim_{h \to 0} \frac{\frac{1}{(3+h)^2} - \frac{1}{9}}{h}

  1. −23-\frac{2}{3}
  2. −227-\frac{2}{27} (correct answer)
  3. 227\frac{2}{27}
  4. 66
Explanation: When you see a limit in the form lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, you're looking at the definition of a derivative – the instantaneous rate of change of function f(x)f(x) at point x=ax = a. Here, we can identify f(x)=1x2f(x) = \frac{1}{x^2} and a=3a = 3. To evaluate this limit, we need to simplify the complex fraction. First, let's work with the numerator: 1(3+h)2−19\frac{1}{(3+h)^2} - \frac{1}{9} To subtract these fractions, we need a common denominator of 9(3+h)29(3+h)^2: 9−(3+h)29(3+h)2\frac{9 - (3+h)^2}{9(3+h)^2} Expanding (3+h)2=9+6h+h2(3+h)^2 = 9 + 6h + h^2, so: 9−(9+6h+h2)9(3+h)2=−6h−h29(3+h)2\frac{9 - (9 + 6h + h^2)}{9(3+h)^2} = \frac{-6h - h^2}{9(3+h)^2} Now our limit becomes: lim⁡h→0−6h−h2h⋅9(3+h)2=lim⁡h→0h(−6−h)h⋅9(3+h)2=lim⁡h→0−6−h9(3+h)2\lim_{h \to 0} \frac{-6h - h^2}{h \cdot 9(3+h)^2} = \lim_{h \to 0} \frac{h(-6 - h)}{h \cdot 9(3+h)^2} = \lim_{h \to 0} \frac{-6 - h}{9(3+h)^2} Substituting h=0h = 0: −6−09(3+0)2=−69⋅9=−681=−227\frac{-6 - 0}{9(3+0)^2} = \frac{-6}{9 \cdot 9} = \frac{-6}{81} = -\frac{2}{27} This confirms answer (B). (A) −23-\frac{2}{3} likely comes from incorrectly simplifying the denominator. (C) 227\frac{2}{27} has the wrong sign – a common error when expanding (3+h)2(3+h)^2. (D) 66 ignores the denominator entirely. Study tip: When evaluating derivative limits, always factor out hh from the numerator before canceling. This prevents the indeterminate form 00\frac{0}{0}.

Question 9

Find the value of the limit: lim⁡x→4x+5−3x−4\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{x-4}

  1. 66
  2. 00
  3. 13\frac{1}{3}
  4. 16\frac{1}{6} (correct answer)
Explanation: When you encounter a limit that produces the indeterminate form 00\frac{0}{0}, you need to use algebraic techniques to resolve it. Let's check: as x→4x \to 4, the numerator becomes 4+5−3=9−3=3−3=0\sqrt{4+5} - 3 = \sqrt{9} - 3 = 3 - 3 = 0, and the denominator becomes 4−4=04 - 4 = 0. This confirms we have the indeterminate form 00\frac{0}{0}. The key technique here is rationalizing the numerator by multiplying both the numerator and denominator by the conjugate x+5+3\sqrt{x+5} + 3: lim⁡x→4x+5−3x−4⋅x+5+3x+5+3\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{x-4} \cdot \frac{\sqrt{x+5} + 3}{\sqrt{x+5} + 3} This gives us: lim⁡x→4(x+5)2−9(x−4)(x+5+3)=lim⁡x→4x+5−9(x−4)(x+5+3)=lim⁡x→4x−4(x−4)(x+5+3)\lim_{x \to 4} \frac{(\sqrt{x+5})^2 - 9}{(x-4)(\sqrt{x+5} + 3)} = \lim_{x \to 4} \frac{x+5-9}{(x-4)(\sqrt{x+5} + 3)} = \lim_{x \to 4} \frac{x-4}{(x-4)(\sqrt{x+5} + 3)} The (x−4)(x-4) terms cancel, leaving: lim⁡x→41x+5+3=14+5+3=19+3=13+3=16\lim_{x \to 4} \frac{1}{\sqrt{x+5} + 3} = \frac{1}{\sqrt{4+5} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} Choice A) 66 represents taking the reciprocal incorrectly. Choice B) 00 might come from incorrectly concluding the limit equals the indeterminate form. Choice C) 13\frac{1}{3} results from forgetting to add the 3 in the denominator after substitution. Remember: when you see a square root in the numerator creating an indeterminate form, try rationalizing with the conjugate. This technique frequently appears in business calculus limit problems.

Question 10

Evaluate the limit: lim⁡x→∞(x2+6x−x)\lim_{x \to \infty} \left( \sqrt{x^2 + 6x} - x \right)

  1. 00
  2. 66
  3. 33 (correct answer)
  4. ∞\infty
Explanation: When you encounter a limit involving the difference of two expressions that both approach infinity, you're dealing with an indeterminate form (∞ - ∞). The key strategy is to rationalize by multiplying by the conjugate. To evaluate lim⁡x→∞(x2+6x−x)\lim_{x \to \infty} \left( \sqrt{x^2 + 6x} - x \right), multiply by the conjugate x2+6x+xx2+6x+x\frac{\sqrt{x^2 + 6x} + x}{\sqrt{x^2 + 6x} + x}: lim⁡x→∞(x2+6x−x)⋅x2+6x+xx2+6x+x\lim_{x \to \infty} \left( \sqrt{x^2 + 6x} - x \right) \cdot \frac{\sqrt{x^2 + 6x} + x}{\sqrt{x^2 + 6x} + x} This gives us: lim⁡x→∞(x2+6x)−x2x2+6x+x=lim⁡x→∞6xx2+6x+x\lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{6x}{\sqrt{x^2 + 6x} + x} Now divide numerator and denominator by xx: lim⁡x→∞61+6x+1\lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1} As x→∞x \to \infty, 6x→0\frac{6}{x} \to 0, so this becomes: 61+0+1=61+1=62=3\frac{6}{\sqrt{1 + 0} + 1} = \frac{6}{1 + 1} = \frac{6}{2} = 3 Answer A) 00 would result from incorrectly thinking both terms cancel completely. Answer B) 66 comes from forgetting to add the xx term in the denominator after rationalization. Answer D) ∞\infty reflects the initial misconception that this limit doesn't exist because you have ∞ - ∞. Study tip: Whenever you see limits involving expression−something\sqrt{expression} - something as x→∞x \to \infty, immediately think "rationalize with the conjugate." This technique converts the indeterminate form into something manageable by eliminating the troublesome square root difference.

Question 11

A company's profit function is P(x)=x3−8x2+16xx2−4xP(x) = \frac{x^3 - 8x^2 + 16x}{x^2 - 4x} where xx represents the number of units produced (in thousands). What is lim⁡x→4P(x)\lim_{x \to 4} P(x)?

  1. 00 (correct answer)
  2. 43\frac{4}{3}
  3. 44
  4. The limit does not exist because the function is undefined at x=4x = 4
Explanation: First, factor the numerator and denominator. The numerator x3−8x2+16x=x(x2−8x+16)=x(x−4)2x^3 - 8x^2 + 16x = x(x^2 - 8x + 16) = x(x-4)^2. The denominator x2−4x=x(x−4)x^2 - 4x = x(x-4). So P(x)=x(x−4)2x(x−4)=(x−4)2x−4=x−4P(x) = \frac{x(x-4)^2}{x(x-4)} = \frac{(x-4)^2}{x-4} = x-4 for x≠0,4x \neq 0, 4. Therefore, lim⁡x→4P(x)=lim⁡x→4(x−4)=0\lim_{x \to 4} P(x) = \lim_{x \to 4} (x-4) = 0. Choice B results from incorrectly simplifying the rational function. Choice C assumes the limit equals the input value. Choice D incorrectly concludes that undefined points prevent limits from existing.

Question 12

What is lim⁡h→04+h−2h\lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h}?

  1. 12\frac{1}{2}
  2. 14\frac{1}{4} (correct answer)
  3. 22
  4. 44
Explanation: This limit represents the derivative of f(x)=xf(x) = \sqrt{x} at x=4x = 4, which you can recognize from the standard derivative definition: f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. When you see this indeterminate form 00\frac{0}{0}, you need to algebraically manipulate it before taking the limit. The key technique here is rationalizing the numerator by multiplying both numerator and denominator by the conjugate 4+h+2\sqrt{4+h} + 2: 4+h−2h⋅4+h+24+h+2=(4+h)2−22h(4+h+2)=4+h−4h(4+h+2)=hh(4+h+2)\frac{\sqrt{4+h} - 2}{h} \cdot \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2} = \frac{(\sqrt{4+h})^2 - 2^2}{h(\sqrt{4+h} + 2)} = \frac{4+h-4}{h(\sqrt{4+h} + 2)} = \frac{h}{h(\sqrt{4+h} + 2)} After canceling the hh terms: 14+h+2\frac{1}{\sqrt{4+h} + 2} Now taking the limit as h→0h \to 0: 14+2=14\frac{1}{\sqrt{4} + 2} = \frac{1}{4}, which is answer B. Answer A (12\frac{1}{2}) likely comes from incorrectly thinking the derivative of x\sqrt{x} at x=4x=4 is 124=14\frac{1}{2\sqrt{4}} = \frac{1}{4} but making an arithmetic error. Answer C (22) might result from confusing this with f(4)=4=2f(4) = \sqrt{4} = 2 rather than f′(4)f'(4). Answer D (44) could come from misremembering the derivative formula entirely. Study tip: When you see limits in the form f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h}, recognize these as derivative definitions. For square root functions, always try rationalizing the numerator with the conjugate to eliminate the indeterminate form.

Question 13

For the function f(x)=x+9−3xf(x) = \frac{\sqrt{x+9} - 3}{x}, what is lim⁡x→0f(x)\lim_{x \to 0} f(x)?

  1. 00
  2. 13\frac{1}{3}
  3. 16\frac{1}{6} (correct answer)
  4. The limit does not exist due to division by zero
Explanation: When you encounter a limit that results in the indeterminate form 00\frac{0}{0}, you need to algebraically manipulate the expression before evaluating. This function gives us 00\frac{0}{0} when we substitute x=0x = 0 directly, so we must rationalize the numerator. To find this limit, multiply both numerator and denominator by the conjugate x+9+3\sqrt{x+9} + 3: lim⁡x→0x+9−3x⋅x+9+3x+9+3\lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x} \cdot \frac{\sqrt{x+9} + 3}{\sqrt{x+9} + 3} The numerator becomes (x+9)2−32=(x+9)−9=x(\sqrt{x+9})^2 - 3^2 = (x+9) - 9 = x. So we have: lim⁡x→0xx(x+9+3)=lim⁡x→01x+9+3\lim_{x \to 0} \frac{x}{x(\sqrt{x+9} + 3)} = \lim_{x \to 0} \frac{1}{\sqrt{x+9} + 3} Now we can safely substitute x=0x = 0: 10+9+3=19+3=13+3=16\frac{1}{\sqrt{0+9} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6} Answer choice (A) 00 is incorrect—this would occur if the numerator approached zero faster than the denominator. Choice (B) 13\frac{1}{3} is wrong; this results from forgetting to add 3 in the denominator after rationalization. Choice (D) is incorrect because while direct substitution gives division by zero, the limit still exists after proper algebraic manipulation. Study tip: When you see square roots in rational functions that produce 00\frac{0}{0}, immediately think "rationalize." Multiply by the conjugate to eliminate the indeterminate form, then evaluate the simplified expression.

Question 14

Consider the limit lim⁡h→0(3+h)2−9h\lim_{h \to 0} \frac{(3+h)^2 - 9}{h}. This expression represents the derivative of f(x)=x2f(x) = x^2 at which point, and what is the value of the limit?

  1. At x=9x = 9; the limit equals 33
  2. At x=0x = 0; the limit equals 99
  3. At x=3x = 3; the limit equals 66 (correct answer)
  4. At x=3x = 3; the limit equals 99
Explanation: When you encounter a limit expression in the form lim⁡h→0f(a+h)−f(a)h\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, you're looking at the definition of a derivative at a specific point. This is the fundamental building block for understanding how calculus measures instantaneous rates of change. Let's identify what's happening here. The expression (3+h)2−9h\frac{(3+h)^2 - 9}{h} follows the derivative definition pattern where f(x)=x2f(x) = x^2, and we're evaluating at x=3x = 3 since (3+h)2(3+h)^2 suggests a=3a = 3. Notice that f(3)=32=9f(3) = 3^2 = 9, which matches the subtracted constant. To find the limit value, expand (3+h)2=9+6h+h2(3+h)^2 = 9 + 6h + h^2. Substituting: 9+6h+h2−9h=6h+h2h=h(6+h)h=6+h\frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = \frac{h(6 + h)}{h} = 6 + h. As h→0h \to 0, this approaches 66. Therefore, this represents the derivative of f(x)=x2f(x) = x^2 at x=3x = 3, and the limit equals 66. Choice A incorrectly identifies the point as x=9x = 9 (confusing the function value with the input) and gets the wrong limit. Choice B misidentifies the point as x=0x = 0 and calculates an incorrect limit of 99. Choice D correctly identifies x=3x = 3 but incorrectly states the limit as 99 instead of 66. Strategy tip: Always match the pattern f(a+h)−f(a)h\frac{f(a+h) - f(a)}{h} to identify both the function and the specific point. The value inside the parentheses with hh tells you the point of evaluation.

Question 15

What is lim⁡x→1x4−1x3−1\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}?

  1. 11
  2. 43\frac{4}{3} (correct answer)
  3. 34\frac{3}{4}
  4. 00
Explanation: When you encounter a limit that produces the indeterminate form 00\frac{0}{0}, you need to manipulate the expression algebraically before evaluating. Let's check: substituting x=1x = 1 gives us 1−11−1=00\frac{1-1}{1-1} = \frac{0}{0}, so we need another approach. The key insight is to factor both the numerator and denominator. For the numerator: x4−1=(x2)2−12=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)x^4 - 1 = (x^2)^2 - 1^2 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1). For the denominator: x3−1=(x−1)(x2+x+1)x^3 - 1 = (x-1)(x^2+x+1) using the difference of cubes formula. Now we can rewrite the limit: lim⁡x→1(x−1)(x+1)(x2+1)(x−1)(x2+x+1)\lim_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{(x-1)(x^2+x+1)} Since we're taking the limit as xx approaches 1 (not equals 1), we can cancel the common factor (x−1)(x-1): lim⁡x→1(x+1)(x2+1)x2+x+1\lim_{x \to 1} \frac{(x+1)(x^2+1)}{x^2+x+1} Now substituting x=1x = 1: (1+1)(12+1)12+1+1=2⋅23=43\frac{(1+1)(1^2+1)}{1^2+1+1} = \frac{2 \cdot 2}{3} = \frac{4}{3} Choice A (11) might tempt you if you incorrectly assume the limit equals 1 because the variable approaches 1. Choice C (34\frac{3}{4}) could result from inverting the correct answer. Choice D (00) might seem right if you mistakenly think the 00\frac{0}{0} form means the limit is 0. Study tip: When you see 00\frac{0}{0}, always factor first. Look for common factors to cancel, especially (x−a)(x-a) where aa is the value the variable approaches.