Calculus 1 Quiz: Limits Via Algebraic Manipulation
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Limits Via Algebraic ManipulationQuestion 1 of 20

Let aa be a non-zero constant. Evaluate lim⁡x→ax2−(a+1)x+ax3−a3\lim_{x\to a} \frac{x^2 - (a+1)x + a}{x^3 - a^3}

13a\frac{1}{3a}
a+13a2\frac{a+1}{3a^2}
a−13a2\frac{a-1}{3a^2}
1a2\frac{1}{a^2}
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Calculus 1 Quiz

Calculus 1 Quiz: Limits Via Algebraic Manipulation

Practice Limits Via Algebraic Manipulation in Calculus 1 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 Limits Via Algebraic Manipulation, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

Let aa be a non-zero constant. Evaluate lim⁡x→ax2−(a+1)x+ax3−a3\lim_{x\to a} \frac{x^2 - (a+1)x + a}{x^3 - a^3}

  1. 13a\frac{1}{3a}
  2. a+13a2\frac{a+1}{3a^2}
  3. a−13a2\frac{a-1}{3a^2} (correct answer)
  4. 1a2\frac{1}{a^2}
Explanation: When you encounter a limit that produces the indeterminate form 00\frac{0}{0}, you need to factor and simplify before evaluating. Let's substitute x=ax = a into both numerator and denominator to confirm we get 00\frac{0}{0}. For the numerator x2−(a+1)x+ax^2 - (a+1)x + a, we need to factor this quadratic. Since it equals zero when x=ax = a, we know (x−a)(x-a) is a factor. Using polynomial division or factoring: x2−(a+1)x+a=(x−a)(x−1)x^2 - (a+1)x + a = (x-a)(x-1). For the denominator x3−a3x^3 - a^3, we use the difference of cubes formula: x3−a3=(x−a)(x2+ax+a2)x^3 - a^3 = (x-a)(x^2 + ax + a^2). Now our limit becomes: lim⁡x→a(x−a)(x−1)(x−a)(x2+ax+a2)=lim⁡x→ax−1x2+ax+a2\lim_{x\to a} \frac{(x-a)(x-1)}{(x-a)(x^2 + ax + a^2)} = \lim_{x\to a} \frac{x-1}{x^2 + ax + a^2} Since we've removed the common factor (x−a)(x-a), we can now substitute x=ax = a: a−1a2+a⋅a+a2=a−13a2\frac{a-1}{a^2 + a \cdot a + a^2} = \frac{a-1}{3a^2} This confirms answer C is correct. Looking at the wrong answers: A) 13a\frac{1}{3a} ignores the numerator's contribution entirely. B) a+13a2\frac{a+1}{3a^2} incorrectly uses (a+1)(a+1) instead of (a−1)(a-1) in the numerator—a sign error from the factoring. D) 1a2\frac{1}{a^2} has the wrong denominator, suggesting incorrect handling of the cubic term. Strategy tip: Always factor completely when you get 00\frac{0}{0}. Check your factoring by expanding back, and remember that x3−a3=(x−a)(x2+ax+a2)x^3 - a^3 = (x-a)(x^2 + ax + a^2).

Question 2

Evaluate the limit: $$ \lim_{x \to -2} \frac{x^3 + 8}{x^2 + 5x + 6}

  1. 4
  2. 12 (correct answer)
  3. -12
  4. The limit does not exist.
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. We can factor the numerator as a sum of cubes and the denominator as a quadratic. Numerator: x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x+2)(x^2 - 2x + 4) Denominator: x2+5x+6=(x+2)(x+3)x^2 + 5x + 6 = (x+2)(x+3) The expression becomes: $$ \lim_{x \to -2} \frac{(x+2)(x^2 - 2x + 4)}{(x+2)(x+3)} Cancel the $(x+2)$ term: \lim_{x \to -2} \frac{x^2 - 2x + 4}{x+3} Now, substitute $x = -2$: \frac{(-2)^2 - 2(-2) + 4}{-2+3} = \frac{4 + 4 + 4}{1} = 12 Distractor A results from a sign error in the sum of cubes formula, using $x^2+2x+4$. Distractor C results from a sign error in the denominator calculation.

Question 3

Evaluate the limit: $$ \lim_{x \to 3} \frac{\frac{1}{x+1} - \frac{1}{4}}{x-3}

  1. -1/16 (correct answer)
  2. 1/16
  3. -1
  4. 0
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. First, simplify the complex fraction by finding a common denominator for the terms in the numerator. 4−(x+1)4(x+1)x−3=3−x4(x+1)x−3\frac{\frac{4 - (x+1)}{4(x+1)}}{x-3} = \frac{\frac{3-x}{4(x+1)}}{x-3} This can be rewritten as: 3−x4(x+1)(x−3)=−(x−3)4(x+1)(x−3)\frac{3-x}{4(x+1)(x-3)} = \frac{-(x-3)}{4(x+1)(x-3)} Now, the limit becomes: lim⁡x→3−(x−3)4(x+1)(x−3)\lim_{x \to 3} \frac{-(x-3)}{4(x+1)(x-3)} Cancel the (x−3)(x-3) terms: lim⁡x→3−14(x+1)\lim_{x \to 3} \frac{-1}{4(x+1)} Substitute x=3x = 3: −14(3+1)=−116\frac{-1}{4(3+1)} = \frac{-1}{16} Distractor B is a sign error. Distractor C results from forgetting the denominator 4(x+1)4(x+1) when simplifying the numerator.

Question 4

Evaluate the limit: lim⁡x→1x4−1x3−1\lim_{x \to 1} \frac{x^4 - 1}{x^3 - 1}

  1. 1
  2. 4/3 (correct answer)
  3. 3/4
  4. 2/3
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. We factor the numerator and the denominator. Numerator (difference of squares twice): x4−1=(x2−1)(x2+1)=(x−1)(x+1)(x2+1)x^4 - 1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1). Denominator (difference of cubes): x3−1=(x−1)(x2+x+1)x^3 - 1 = (x-1)(x^2+x+1). The limit becomes: 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)} Cancel the (x−1)(x-1) term: lim⁡x→1(x+1)(x2+1)x2+x+1\lim_{x \to 1} \frac{(x+1)(x^2+1)}{x^2+x+1} Substitute 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} Distractor D might result from an incorrect factorization. Distractor C is the reciprocal of the answer.

Question 5

For the limit lim⁡x→−1x2+cx−2x+1\lim_{x \to -1} \frac{x^2 + cx - 2}{x+1} to exist, what must be the value of the constant cc?

  1. -2
  2. -1 (correct answer)
  3. 1
  4. 2
Explanation: For a limit of a rational function to exist where the denominator approaches zero, the numerator must also approach zero. This allows for the possibility of canceling the problematic factor. Set the numerator equal to zero at x=−1x = -1: (−1)2+c(−1)−2=0(-1)^2 + c(-1) - 2 = 0 1−c−2=01 - c - 2 = 0 −1−c=0-1 - c = 0 c=−1c = -1 With c=−1c=-1, the limit becomes lim⁡x→−1x2−x−2x+1=lim⁡x→−1(x−2)(x+1)x+1=−3\lim_{x \to -1} \frac{x^2 - x - 2}{x+1} = \lim_{x \to -1} \frac{(x-2)(x+1)}{x+1} = -3, which exists. Distractor C is a sign error in solving for cc.

Question 6

Evaluate the limit: lim⁡x→−2x3+2x2−3x−6x2−x−6\lim_{x\to -2} \frac{x^3 + 2x^2 - 3x - 6}{x^2 - x - 6}

  1. 1/51/5
  2. −1/5-1/5 (correct answer)
  3. −1-1
  4. 11
Explanation: When you encounter a rational function limit where direct substitution gives the indeterminate form 00\frac{0}{0}, you need to factor and simplify first. Let's check: substituting x=−2x = -2 into both numerator and denominator gives 00\frac{0}{0}, so we must factor. For the numerator x3+2x2−3x−6x^3 + 2x^2 - 3x - 6, try factoring by grouping: x2(x+2)−3(x+2)=(x2−3)(x+2)x^2(x + 2) - 3(x + 2) = (x^2 - 3)(x + 2). The denominator x2−x−6x^2 - x - 6 factors as (x−3)(x+2)(x - 3)(x + 2) since we need two numbers that multiply to −6-6 and add to −1-1. Now we can simplify: (x2−3)(x+2)(x−3)(x+2)=x2−3x−3\frac{(x^2 - 3)(x + 2)}{(x - 3)(x + 2)} = \frac{x^2 - 3}{x - 3} after canceling the common factor (x+2)(x + 2). Substituting x=−2x = -2 into the simplified expression: (−2)2−3(−2)−3=4−3−5=1−5=−15\frac{(-2)^2 - 3}{(-2) - 3} = \frac{4 - 3}{-5} = \frac{1}{-5} = -\frac{1}{5} Answer choice A gives 15\frac{1}{5}, which would result from a sign error in the denominator calculation. Choice C gives −1-1, likely from incorrectly simplifying 1−5\frac{1}{-5} or making an arithmetic mistake. Choice D gives 1, which combines both the sign error and the arithmetic error. The correct answer is B: −15-\frac{1}{5}. Key strategy: When limits give 00\frac{0}{0}, always factor completely and cancel common factors before substituting. Double-check your arithmetic, especially with negative signs, as these are common sources of errors on calculus exams.

Question 7

Evaluate the limit: lim⁡x→−21x+12x3+8\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x^3+8}

  1. 1/12
  2. -1/16
  3. -1/48 (correct answer)
  4. 1/24
Explanation: This limit is of the indeterminate form 00\frac{0}{0}. First, simplify the complex fraction in the numerator. 1x+12=2+x2x\frac{1}{x} + \frac{1}{2} = \frac{2+x}{2x} The expression becomes: lim⁡x→−2x+22xx3+8\lim_{x \to -2} \frac{\frac{x+2}{2x}}{x^3+8} Factor the denominator as a sum of cubes: x3+8=(x+2)(x2−2x+4)x^3+8 = (x+2)(x^2-2x+4). lim⁡x→−2x+22x(x+2)(x2−2x+4)\lim_{x \to -2} \frac{x+2}{2x(x+2)(x^2-2x+4)} Cancel the (x+2)(x+2) term: lim⁡x→−212x(x2−2x+4)\lim_{x \to -2} \frac{1}{2x(x^2-2x+4)} Substitute x=−2x=-2: 12(−2)((−2)2−2(−2)+4)=1−4(4+4+4)=1−4(12)=−148\frac{1}{2(-2)((-2)^2-2(-2)+4)} = \frac{1}{-4(4+4+4)} = \frac{1}{-4(12)} = -\frac{1}{48} Distractor A results from forgetting the 2x2x term from the initial simplification. Distractor B results from a sign error in the sum of cubes formula.

Question 8

Evaluate the limit: $$ \lim_{x \to 1} \frac{(\sqrt{x}-1)^2}{x-1}

  1. 0 (correct answer)
  2. 1
  3. 1/2
  4. The limit does not exist.
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. The key is to factor the denominator as a difference of squares: x−1=(x−1)(x+1)x-1 = (\sqrt{x}-1)(\sqrt{x}+1). The expression becomes: lim⁡x→1(x−1)2(x−1)(x+1)\lim_{x \to 1} \frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)} Cancel one factor of (x−1)(\sqrt{x}-1) from the numerator and denominator: lim⁡x→1x−1x+1\lim_{x \to 1} \frac{\sqrt{x}-1}{\sqrt{x}+1} Now, substitute x=1x=1: 1−11+1=1−11+1=02=0\frac{\sqrt{1}-1}{\sqrt{1}+1} = \frac{1-1}{1+1} = \frac{0}{2} = 0 Distractor C results from incorrectly canceling the numerator, leaving 1 instead of (x−1)(\sqrt{x}-1).

Question 9

Evaluate the limit: lim⁡x→1x1/3−1x1/4−1\lim_{x\to 1} \frac{x^{1/3} - 1}{x^{1/4} - 1}

  1. 11
  2. 3/43/4
  3. 4/34/3 (correct answer)
  4. 00
Explanation: When you encounter a limit that produces the indeterminate form 00\frac{0}{0}, you need a strategy to resolve it. This limit gives 1−11−1=00\frac{1-1}{1-1} = \frac{0}{0} when you substitute x=1x = 1 directly. The most efficient approach here is to use the substitution u=x1/12u = x^{1/12}, which means x=u12x = u^{12}. As x→1x \to 1, we have u→1u \to 1. This transforms our expression into: x1/3−1x1/4−1=(u12)1/3−1(u12)1/4−1=u4−1u3−1\frac{x^{1/3} - 1}{x^{1/4} - 1} = \frac{(u^{12})^{1/3} - 1}{(u^{12})^{1/4} - 1} = \frac{u^4 - 1}{u^3 - 1} Now factor both numerator and denominator:
  • u4−1=(u−1)(u3+u2+u+1)u^4 - 1 = (u-1)(u^3 + u^2 + u + 1)
  • u3−1=(u−1)(u2+u+1)u^3 - 1 = (u-1)(u^2 + u + 1)
After canceling the common factor (u−1)(u-1): u4−1u3−1=u3+u2+u+1u2+u+1\frac{u^4 - 1}{u^3 - 1} = \frac{u^3 + u^2 + u + 1}{u^2 + u + 1} Taking the limit as u→1u \to 1: 1+1+1+11+1+1=43\frac{1 + 1 + 1 + 1}{1 + 1 + 1} = \frac{4}{3} So the answer is C) 4/34/3. Choice A) 1 might result from incorrectly assuming the powers don't matter. Choice B) 3/43/4 comes from flipping the fraction of exponents 1/31/4\frac{1/3}{1/4}. Choice D) 0 suggests misunderstanding indeterminate forms. Key strategy: When facing 00\frac{0}{0} limits with fractional exponents, find a common denominator for the exponents and use substitution to convert to polynomial forms that factor easily.

Question 10

Evaluate the limit: lim⁡x→0sec⁡(x)−1xsec⁡(x)\lim_{x\to 0} \frac{\sec(x) - 1}{x \sec(x)}

  1. 11
  2. 00 (correct answer)
  3. 1/21/2
  4. Does not exist
Explanation: When you encounter a limit that produces an indeterminate form like 00\frac{0}{0}, you need to manipulate the expression algebraically before applying limit properties. Let's check: as x→0x \to 0, we have sec⁡(0)=1\sec(0) = 1, so the numerator approaches 1−1=01 - 1 = 0 and the denominator approaches 0⋅1=00 \cdot 1 = 0. To resolve this indeterminate form, rewrite sec⁡(x)=1cos⁡(x)\sec(x) = \frac{1}{\cos(x)}: lim⁡x→0sec⁡(x)−1xsec⁡(x)=lim⁡x→01cos⁡(x)−1x⋅1cos⁡(x)\lim_{x\to 0} \frac{\sec(x) - 1}{x \sec(x)} = \lim_{x\to 0} \frac{\frac{1}{\cos(x)} - 1}{x \cdot \frac{1}{\cos(x)}} Simplify the numerator: 1cos⁡(x)−1=1−cos⁡(x)cos⁡(x)\frac{1}{\cos(x)} - 1 = \frac{1 - \cos(x)}{\cos(x)} The expression becomes: lim⁡x→01−cos⁡(x)cos⁡(x)xcos⁡(x)=lim⁡x→01−cos⁡(x)x\lim_{x\to 0} \frac{\frac{1 - \cos(x)}{\cos(x)}}{\frac{x}{\cos(x)}} = \lim_{x\to 0} \frac{1 - \cos(x)}{x} This is a standard limit! Using the fundamental limit lim⁡x→01−cos⁡(x)x=0\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0 (which you can verify using L'Hôpital's rule or the identity 1−cos⁡(x)=2sin⁡2(x/2)1 - \cos(x) = 2\sin^2(x/2)), we get 0. Choice A (11) might tempt you if you incorrectly applied L'Hôpital's rule to the original form. Choice C (1/21/2) could result from confusing this with the related limit lim⁡x→01−cos⁡(x)x2=12\lim_{x\to 0} \frac{1 - \cos(x)}{x^2} = \frac{1}{2}. Choice D is wrong because the limit clearly exists after proper algebraic manipulation. Study tip: Master the standard trigonometric limits like lim⁡x→0sin⁡(x)x=1\lim_{x\to 0} \frac{\sin(x)}{x} = 1 and lim⁡x→01−cos⁡(x)x=0\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0. When facing indeterminate forms, always simplify algebraically first before applying these fundamental results.

Question 11

Let ff be the function defined by f(x)={x−2x−1−1if x>2ax+1if x≤2f(x) = \begin{cases} \frac{x-2}{\sqrt{x-1}-1} & \text{if } x > 2 \\ ax + 1 & \text{if } x \le 2 \end{cases}. For what value of the constant aa is the function ff continuous at x=2x = 2?

  1. 11
  2. 22
  3. 1/21/2 (correct answer)
  4. −1/2-1/2
Explanation: When you encounter a piecewise function and need to find a constant that makes it continuous at a boundary point, you're testing whether the left-hand and right-hand limits equal the function value at that point. For continuity at x=2x = 2, we need lim⁡x→2−f(x)=lim⁡x→2+f(x)=f(2)\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2). First, find the left-hand limit. Since x≤2x \leq 2 uses f(x)=ax+1f(x) = ax + 1, we have lim⁡x→2−f(x)=a(2)+1=2a+1\lim_{x \to 2^-} f(x) = a(2) + 1 = 2a + 1. Also, f(2)=2a+1f(2) = 2a + 1. Next, find the right-hand limit using f(x)=x−2x−1−1f(x) = \frac{x-2}{\sqrt{x-1}-1} for x>2x > 2. This gives the indeterminate form 00\frac{0}{0} as x→2+x \to 2^+, so we need to simplify. Multiply by the conjugate: x−2x−1−1⋅x−1+1x−1+1=(x−2)(x−1+1)(x−1)−1=(x−2)(x−1+1)x−2=x−1+1\frac{x-2}{\sqrt{x-1}-1} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1} = \frac{(x-2)(\sqrt{x-1}+1)}{(x-1)-1} = \frac{(x-2)(\sqrt{x-1}+1)}{x-2} = \sqrt{x-1}+1 Therefore, lim⁡x→2+f(x)=2−1+1=1+1=2\lim_{x \to 2^+} f(x) = \sqrt{2-1}+1 = 1+1 = 2. Setting the limits equal: 2a+1=22a + 1 = 2, so a=12a = \frac{1}{2}. Looking at the wrong answers: A) a=1a = 1 gives 2a+1=3≠22a + 1 = 3 \neq 2. B) a=2a = 2 gives 2a+1=5≠22a + 1 = 5 \neq 2. D) a=−12a = -\frac{1}{2} gives 2a+1=0≠22a + 1 = 0 \neq 2. Study tip: For piecewise continuity problems, always rationalize expressions with square roots when you encounter 00\frac{0}{0} forms—the conjugate method is your key tool.

Question 12

Let f(x)=ax2+bx−4x−2f(x) = \frac{ax^2+bx-4}{x-2}. If lim⁡x→2f(x)=6\lim_{x \to 2} f(x) = 6, what is the value of bb?

  1. -2 (correct answer)
  2. 2
  3. 0
  4. -4
Explanation: For the limit to exist as x→2x \to 2, the numerator must be zero at x=2x=2 to cancel the zero in the denominator. So, a(2)2+b(2)−4=0a(2)^2 + b(2) - 4 = 0, which simplifies to 4a+2b=44a + 2b = 4, or 2a+b=22a+b=2. From this, b=2−2ab = 2-2a. Substitute this expression for bb into the numerator: ax2+(2−2a)x−4ax^2 + (2-2a)x - 4 This polynomial must have a factor of (x−2)(x-2). Let's factor by grouping: ax2−2ax+2x−4=ax(x−2)+2(x−2)=(ax+2)(x−2)ax^2 - 2ax + 2x - 4 = ax(x-2) + 2(x-2) = (ax+2)(x-2) Now, we can evaluate the limit: lim⁡x→2(ax+2)(x−2)x−2=lim⁡x→2(ax+2)=2a+2\lim_{x \to 2} \frac{(ax+2)(x-2)}{x-2} = \lim_{x \to 2} (ax+2) = 2a+2 We are given that the limit is 6. So, 2a+2=62a+2 = 6, which means 2a=42a = 4 and a=2a = 2. Finally, find bb using the relationship b=2−2ab = 2-2a: b=2−2(2)=2−4=−2b = 2 - 2(2) = 2 - 4 = -2. Distractor B is the value of aa. Distractor C is the value of a+ba+b.

Question 13

Evaluate the limit: $$ \lim_{x \to 0} \frac{1}{x} \left( \frac{1}{2+x} - \frac{1}{2} \right)

  1. -1
  2. 1/4
  3. -1/4 (correct answer)
  4. The limit does not exist.
Explanation: The expression is of the indeterminate form ∞⋅0\infty \cdot 0. First, combine the fractions inside the parentheses. 12+x−12=2−(2+x)2(2+x)=−x2(2+x)\frac{1}{2+x} - \frac{1}{2} = \frac{2 - (2+x)}{2(2+x)} = \frac{-x}{2(2+x)} Now substitute this back into the limit expression: lim⁡x→01x(−x2(2+x))\lim_{x \to 0} \frac{1}{x} \left( \frac{-x}{2(2+x)} \right) Cancel the xx terms: lim⁡x→0−12(2+x)\lim_{x \to 0} \frac{-1}{2(2+x)} Substitute x=0x=0: −12(2+0)=−14\frac{-1}{2(2+0)} = -\frac{1}{4} Distractor A results from forgetting the denominator 2(2+x)2(2+x) after simplification. Distractor B is a sign error.

Question 14

Evaluate the limit: $$ \lim_{x \to -\infty} \frac{|x^2-5|}{1-2x^2}

  1. 1/2
  2. -1/2 (correct answer)
  3. -5
  4. \infty
Explanation: As x→−∞x \to -\infty, x2x^2 becomes a very large positive number. Thus, x2−5x^2-5 is positive. Therefore, ∣x2−5∣=x2−5|x^2-5| = x^2-5 for large negative xx. The limit becomes: lim⁡x→−∞x2−51−2x2\lim_{x \to -\infty} \frac{x^2-5}{1-2x^2} To evaluate this limit at infinity, we can take the ratio of the coefficients of the highest power terms (x2x^2) in the numerator and denominator. 1−2=−12\frac{1}{-2} = -\frac{1}{2} Alternatively, divide every term by x2x^2: lim⁡x→−∞1−5x21x2−2=1−00−2=−12\lim_{x \to -\infty} \frac{1 - \frac{5}{x^2}}{\frac{1}{x^2}-2} = \frac{1-0}{0-2} = -\frac{1}{2} Distractor A is a common error from assuming that the absolute value makes the entire result positive, or from incorrectly evaluating ∣x2−5∣=−(x2−5)|x^2-5| = -(x^2-5).

Question 15

Evaluate the limit: $$ \lim_{x \to 1} \frac{x-1}{\sqrt{x^2+3}-2}

  1. 1/2
  2. 2 (correct answer)
  3. 4
  4. The limit does not exist.
Explanation: The limit is of the indeterminate form 00\frac{0}{0}. Multiply the numerator and denominator by the conjugate of the denominator, x2+3+2\sqrt{x^2+3}+2. lim⁡x→1x−1x2+3−2⋅x2+3+2x2+3+2\lim_{x \to 1} \frac{x-1}{\sqrt{x^2+3}-2} \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} =lim⁡x→1(x−1)(x2+3+2)(x2+3)−4= \lim_{x \to 1} \frac{(x-1)(\sqrt{x^2+3}+2)}{(x^2+3)-4} =lim⁡x→1(x−1)(x2+3+2)x2−1= \lim_{x \to 1} \frac{(x-1)(\sqrt{x^2+3}+2)}{x^2-1} Factor the denominator x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1): =lim⁡x→1(x−1)(x2+3+2)(x−1)(x+1)= \lim_{x \to 1} \frac{(x-1)(\sqrt{x^2+3}+2)}{(x-1)(x+1)} Cancel the (x−1)(x-1) term: =lim⁡x→1x2+3+2x+1= \lim_{x \to 1} \frac{\sqrt{x^2+3}+2}{x+1} Substitute x=1x=1: 12+3+21+1=4+22=2+22=2\frac{\sqrt{1^2+3}+2}{1+1} = \frac{\sqrt{4}+2}{2} = \frac{2+2}{2} = 2 Distractor A is a very common error resulting from forgetting to multiply the numerator by the conjugate term.

Question 16

Evaluate the limit: lim⁡x→8x3−2x−8\lim_{x\to 8} \frac{\sqrt[3]{x} - 2}{x - 8}

  1. 1/121/12 (correct answer)
  2. 1/81/8
  3. 1/41/4
  4. 00
Explanation: This limit is in the indeterminate form 0/00/0. We can resolve this by using the algebraic identity for the difference of cubes, a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a-b)(a^2+ab+b^2). Let a=x3a = \sqrt[3]{x} and b=2b = 2. We multiply the numerator and denominator by the 'conjugate' term ((x3)2+2x3+4)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4): lim⁡x→8x3−2x−8⋅(x3)2+2x3+4(x3)2+2x3+4=lim⁡x→8x−8(x−8)((x3)2+2x3+4)\lim_{x\to 8} \frac{\sqrt[3]{x} - 2}{x - 8} \cdot \frac{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4}{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4} = \lim_{x\to 8} \frac{x - 8}{(x - 8)((\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4)} After canceling the (x−8)(x-8) terms, we evaluate the limit: lim⁡x→81(x3)2+2x3+4=1(83)2+283+4=122+2(2)+4=14+4+4=112\lim_{x\to 8} \frac{1}{(\sqrt[3]{x})^2 + 2\sqrt[3]{x} + 4} = \frac{1}{(\sqrt[3]{8})^2 + 2\sqrt[3]{8} + 4} = \frac{1}{2^2 + 2(2) + 4} = \frac{1}{4+4+4} = \frac{1}{12} Distractor B, 1/81/8, might result from an error in evaluating the denominator as 2+2+4=82+2+4=8. Distractor C, 1/41/4, might result from only considering one of the three terms in the denominator. Distractor D, 00, is the result of incorrect direct substitution into the numerator.

Question 17

What is the value of the limit lim⁡h→01(h+3)2−19h\lim_{h\to 0} \frac{\frac{1}{(h+3)^2} - \frac{1}{9}}{h}?

  1. −2/27-2/27 (correct answer)
  2. 2/272/27
  3. −1/27-1/27
  4. 1/811/81
Explanation: This limit is of the form 0/00/0. To evaluate it, we simplify the complex fraction in the numerator by finding a common denominator: 1(h+3)2−19=9−(h+3)29(h+3)2=9−(h2+6h+9)9(h+3)2=−h2−6h9(h+3)2=−h(h+6)9(h+3)2\frac{1}{(h+3)^2} - \frac{1}{9} = \frac{9 - (h+3)^2}{9(h+3)^2} = \frac{9 - (h^2+6h+9)}{9(h+3)^2} = \frac{-h^2-6h}{9(h+3)^2} = \frac{-h(h+6)}{9(h+3)^2} Now, substitute this back into the original limit expression: lim⁡h→0−h(h+6)9(h+3)2h=lim⁡h→0−h(h+6)9h(h+3)2\lim_{h\to 0} \frac{\frac{-h(h+6)}{9(h+3)^2}}{h} = \lim_{h\to 0} \frac{-h(h+6)}{9h(h+3)^2} We can cancel the hh term: lim⁡h→0−(h+6)9(h+3)2\lim_{h\to 0} \frac{-(h+6)}{9(h+3)^2} Now, substitute h=0h=0: −(0+6)9(0+3)2=−69(9)=−681=−227\frac{-(0+6)}{9(0+3)^2} = \frac{-6}{9(9)} = \frac{-6}{81} = -\frac{2}{27} Distractor B reflects a sign error. Distractors C and D result from errors in expanding (h+3)2(h+3)^2 or simplifying the numerator.

Question 18

Evaluate the limit: lim⁡x→π/41−tan⁡(x)cos⁡(x)−sin⁡(x)\lim_{x\to \pi/4} \frac{1 - \tan(x)}{\cos(x) - \sin(x)}

  1. −1-1
  2. −2-\sqrt{2}
  3. 2/2\sqrt{2}/2
  4. 2\sqrt{2} (correct answer)
Explanation: When you encounter a limit that produces an indeterminate form like 00\frac{0}{0}, you need to use algebraic manipulation or L'Hôpital's rule. Let's first check what happens when we substitute x=π/4x = \pi/4: both 1−tan⁡(π/4)=1−1=01 - \tan(\pi/4) = 1 - 1 = 0 and cos⁡(π/4)−sin⁡(π/4)=22−22=0\cos(\pi/4) - \sin(\pi/4) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0, confirming we have the indeterminate form 00\frac{0}{0}. The key insight is to factor the denominator cleverly. Notice that cos⁡(x)−sin⁡(x)=−(sin⁡(x)−cos⁡(x))\cos(x) - \sin(x) = -(\sin(x) - \cos(x)). Now, multiply both numerator and denominator by the conjugate (cos⁡(x)+sin⁡(x))(\cos(x) + \sin(x)): 1−tan⁡(x)cos⁡(x)−sin⁡(x)⋅cos⁡(x)+sin⁡(x)cos⁡(x)+sin⁡(x)=(1−tan⁡(x))(cos⁡(x)+sin⁡(x))cos⁡2(x)−sin⁡2(x)\frac{1 - \tan(x)}{\cos(x) - \sin(x)} \cdot \frac{\cos(x) + \sin(x)}{\cos(x) + \sin(x)} = \frac{(1 - \tan(x))(\cos(x) + \sin(x))}{\cos^2(x) - \sin^2(x)} Since tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)}, the numerator becomes (cos⁡(x)+sin⁡(x))−sin⁡(x)−sin⁡2(x)/cos⁡(x)=(cos⁡(x)+sin⁡(x))−sin⁡(x)(1+sin⁡(x)/cos⁡(x))=cos⁡(x)−sin⁡(x)(\cos(x) + \sin(x)) - \sin(x) - \sin^2(x)/\cos(x) = (\cos(x) + \sin(x)) - \sin(x)(1 + \sin(x)/\cos(x)) = \cos(x) - \sin(x) Wait, let me use a simpler approach. Factor out −1-1 from the denominator: 1−tan⁡(x)−(sin⁡(x)−cos⁡(x))=−(tan(x)−1)sin⁡(x)−cos⁡(x)\frac{1 - \tan(x)}{-(\sin(x) - \cos(x))} = \frac{-(tan(x) - 1)}{\sin(x) - \cos(x)} Using L'Hôpital's rule: lim⁡x→π/4−sec⁡2(x)−sin⁡(x)+cos⁡(x)=−2−22+22=2\lim_{x\to \pi/4} \frac{-\sec^2(x)}{-\sin(x) + \cos(x)} = \frac{-2}{-\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} = \sqrt{2} This confirms answer D is correct. Answer A (-1) ignores the trigonometric values at π/4\pi/4. Answer B (-√2) has the wrong sign. Answer C (√2/2) likely comes from confusing the trigonometric values themselves with the limit result. Study tip: When facing 00\frac{0}{0} limits with trigonometric functions, try algebraic manipulation first, then L'Hôpital's rule as backup.

Question 19

Evaluate the limit: lim⁡x→01x+4−141x+1−1\lim_{x\to 0} \frac{\frac{1}{x+4} - \frac{1}{4}}{\frac{1}{x+1} - 1}

  1. −1/16-1/16
  2. 1/161/16 (correct answer)
  3. 1/41/4
  4. 44
Explanation: This is a limit problem involving complex fractions that initially gives the indeterminate form 0/0. When you encounter limits with fractions in both numerator and denominator, your first step should be to simplify by finding a common denominator for each part. Let's work through this systematically. For the numerator: 1x+4−14=4−(x+4)4(x+4)=−x4(x+4)\frac{1}{x+4} - \frac{1}{4} = \frac{4 - (x+4)}{4(x+4)} = \frac{-x}{4(x+4)} For the denominator: 1x+1−1=1−(x+1)x+1=−xx+1\frac{1}{x+1} - 1 = \frac{1 - (x+1)}{x+1} = \frac{-x}{x+1} Now our limit becomes: lim⁡x→0−x4(x+4)−xx+1=lim⁡x→0−x4(x+4)⋅x+1−x\lim_{x\to 0} \frac{\frac{-x}{4(x+4)}}{\frac{-x}{x+1}} = \lim_{x\to 0} \frac{-x}{4(x+4)} \cdot \frac{x+1}{-x} The −x-x terms cancel (since x≠0x \neq 0 as we approach the limit), giving us: lim⁡x→0x+14(x+4)=0+14(0+4)=116\lim_{x\to 0} \frac{x+1}{4(x+4)} = \frac{0+1}{4(0+4)} = \frac{1}{16} This confirms answer choice B is correct. Let's examine the wrong answers: A) −1/16-1/16 likely results from a sign error during the algebraic manipulation. C) 1/41/4 might come from incorrectly canceling terms or mishandling the denominators. D) 44 could result from flipping the final fraction or major algebraic mistakes. Key strategy: When dealing with complex fraction limits, always simplify each fraction separately first, then combine. Watch your signs carefully throughout the algebra, and remember that common factors in numerator and denominator can often be canceled when approaching (but not at) the limit point.

Question 20

Evaluate the one-sided limit: lim⁡x→3−x2−2x−3∣x−3∣\lim_{x\to 3^-} \frac{x^2 - 2x - 3}{|x-3|}

  1. 22
  2. 44
  3. Does not exist
  4. −4-4 (correct answer)
Explanation: When evaluating one-sided limits involving absolute value expressions, you need to carefully consider what happens to the absolute value on the specified side of the limit point. Since we're approaching 3 from the left (x→3−x \to 3^-), we have x<3x < 3, which means x−3<0x - 3 < 0. Therefore, ∣x−3∣=−(x−3)=3−x|x - 3| = -(x - 3) = 3 - x. Now we can rewrite our limit: lim⁡x→3−x2−2x−33−x\lim_{x\to 3^-} \frac{x^2 - 2x - 3}{3 - x} Next, factor the numerator: x2−2x−3=(x−3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1) Our limit becomes: lim⁡x→3−(x−3)(x+1)3−x=lim⁡x→3−(x−3)(x+1)−(x−3)\lim_{x\to 3^-} \frac{(x - 3)(x + 1)}{3 - x} = \lim_{x\to 3^-} \frac{(x - 3)(x + 1)}{-(x - 3)} We can cancel the (x−3)(x - 3) terms since x≠3x \neq 3 in the limit: lim⁡x→3−x+1−1=lim⁡x→3−−(x+1)\lim_{x\to 3^-} \frac{x + 1}{-1} = \lim_{x\to 3^-} -(x + 1) Evaluating at x=3x = 3: −(3+1)=−4-(3 + 1) = -4 Answer choice A) gives 2, which you'd get if you incorrectly used ∣x−3∣=x−3|x - 3| = x - 3. Answer B) gives 4, which results from the correct factoring but forgetting the negative sign from the absolute value. Answer C) suggests the limit doesn't exist, but one-sided limits can exist even when two-sided limits don't. Remember: always determine what the absolute value equals on your side of the limit before simplifying. The direction of approach determines the sign inside absolute value expressions.