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AP Calculus AB Quiz

AP Calculus AB Quiz: Defining Limits And Using Limit Notation

Practice Defining Limits And Using Limit Notation in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

From the graph, p(x)p(x)p(x) approaches −2-2−2 as xxx approaches 333 from the left. Which limit notation represents this?

Select an answer to continue

What this quiz covers

This quiz focuses on Defining Limits And Using Limit Notation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

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

From the graph, p(x)p(x)p(x) approaches −2-2−2 as xxx approaches 333 from the left. Which limit notation represents this?

  1. lim⁡x→3−p(x)=−2\displaystyle \lim_{x\to 3^-} p(x)=-2x→3−lim​p(x)=−2 (correct answer)
  2. lim⁡x→3+p(x)=−2\displaystyle \lim_{x\to 3^+} p(x)=-2x→3+lim​p(x)=−2
  3. p(3)=−2\displaystyle p(3)=-2p(3)=−2
  4. lim⁡x→−2p(x)=3\displaystyle \lim_{x\to -2} p(x)=3x→−2lim​p(x)=3
  5. lim⁡x→3p(3)=−2\displaystyle \lim_{x\to 3} p(3)=-2x→3lim​p(3)=−2

Explanation: Limit notation highlights approaching values, and the graph shows p(x) nearing -2 from the left as x approaches 3. Hence, \lim_{x\to 3^-} p(x)=-2 is the precise left-hand limit expression. This is valid for capturing behavior from x less than 3. A frequent symbolic mistake is \lim_{x\to 3} p(3)=-2, incorrectly placing the evaluation inside. Using p(3)=-2 confuses limit with function value. Specify one-sided when sides differ. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 2

The graph has an open circle at (2,−3)(2,-3)(2,−3) and the curve approaches −3-3−3 from both sides near x=2x=2x=2; which notation matches?

  1. lim⁡x→2r(x)=−3\displaystyle \lim_{x\to 2} r(x)=-3x→2lim​r(x)=−3 (correct answer)
  2. lim⁡x→−3r(x)=2\displaystyle \lim_{x\to -3} r(x)=2x→−3lim​r(x)=2
  3. lim⁡x→2r(x)=r(2)\displaystyle \lim_{x\to 2} r(x)=r(2)x→2lim​r(x)=r(2)
  4. lim⁡x→2−r(x)=3\displaystyle \lim_{x\to 2^-} r(x)=3x→2−lim​r(x)=3
  5. lim⁡x→2+r(x)=DNE\displaystyle \lim_{x\to 2^+} r(x)=\text{DNE}x→2+lim​r(x)=DNE

Explanation: An open circle on a graph signifies the limit value where the function is not defined or differs, but the curve's approach determines the limit. The graph approaches -3 from both sides at x=2 with an open circle, so \lim_{x \to 2} r(x) = -3 is correct. This notation is valid for the bilateral approach shown. A common symbolic error is assuming DNE due to the hole, ignoring the approaching behavior. Another mistake is using one-sided notation unnecessarily, like \lim_{x \to 2^-} r(x) = 3, or equating to r(2). Verify graph symmetry near the point. Transferable notation checklist: 1. Write as \lim_{x \to a} f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.

Question 3

A function ggg satisfies g(x)=1g(x)=1g(x)=1 for x<3x<3x<3, g(3)=5g(3)=5g(3)=5, and g(x)=4g(x)=4g(x)=4 for x>3x>3x>3. Which limit notation matches ggg near x=3x=3x=3?

  1. lim⁡x→3+g(x)=1\lim_{x\to3^+} g(x)=1limx→3+​g(x)=1
  2. lim⁡x→3g(x)=5\lim_{x\to3} g(x)=5limx→3​g(x)=5
  3. lim⁡x→3−g(x)=4\lim_{x\to3^-} g(x)=4limx→3−​g(x)=4
  4. lim⁡x→3+g(x)=4\lim_{x\to3^+} g(x)=4limx→3+​g(x)=4 (correct answer)
  5. lim⁡x→3g(3)=5\lim_{x\to3} g(3)=5limx→3​g(3)=5

Explanation: The function ggg has a jump discontinuity at x=3x=3x=3: it equals 1 for x<3x<3x<3, jumps to 5 at x=3x=3x=3, then equals 4 for x>3x>3x>3. When approaching from the right (values greater than 3), we're in the region where g(x)=4g(x)=4g(x)=4, so the right-hand limit is lim⁡x→3+g(x)=4\lim_{x\to 3^+} g(x) = 4limx→3+​g(x)=4. The superscript plus sign indicates we only consider values approaching from the right. A common mistake is thinking the limit must involve the function value at the point—but g(3)=5g(3)=5g(3)=5 is irrelevant to the right-hand limit. Another error is writing lim⁡x→3g(3)\lim_{x\to 3} g(3)limx→3​g(3), which incorrectly substitutes the value into the limit notation. Notation checklist: (1) lim⁡\limlim symbol, (2) x→3+x\to 3^+x→3+ for right-hand approach, (3) g(x)g(x)g(x) not g(3)g(3)g(3), (4) equals 4.

Question 4

For the function fff shown in the table, which limit expression represents the behavior of f(x)f(x)f(x) as xxx approaches 222?

  1. lim⁡x→2f(x)=5\displaystyle \lim_{x\to 2} f(x)=5x→2lim​f(x)=5 (correct answer)
  2. f(2)=5\displaystyle f(2)=5f(2)=5
  3. lim⁡x→5f(x)=2\displaystyle \lim_{x\to 5} f(x)=2x→5lim​f(x)=2
  4. lim⁡x→2−f(x)=3\displaystyle \lim_{x\to 2^-} f(x)=3x→2−lim​f(x)=3
  5. lim⁡x→2f(2)=5\displaystyle \lim_{x\to 2} f(2)=5x→2lim​f(2)=5

Explanation: Limit notation is used to describe the value a function approaches as the input nears a certain point, distinct from the function's value at that point. In this scenario, the table indicates that f(x) approaches 5 as x gets close to 2 from both sides, making \lim_{x\to 2} f(x)=5 the appropriate expression. This notation is valid because it focuses on the behavior around x=2 without evaluating f at exactly 2. A common symbolic error is writing \lim_{x\to 2} f(2)=5, which improperly substitutes the point into the function inside the limit. Another frequent mistake is using f(2)=5, which represents the function value, not the limit. Remember, limits can exist even if the function is undefined at the point. Transferable notation checklist: 1. Use 'lim' to denote limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 5

For s(x)=∣x∣xs(x)=\frac{|x|}{x}s(x)=x∣x∣​ when x≠0x\ne0x=0 and s(0)=0s(0)=0s(0)=0, which limit notation describes s(x)s(x)s(x) as x→0x\to0x→0?

  1. lim⁡x→0s(x)=0\displaystyle \lim_{x\to 0} s(x)=0x→0lim​s(x)=0
  2. lim⁡x→0−s(x)=1\displaystyle \lim_{x\to 0^-} s(x)=1x→0−lim​s(x)=1
  3. lim⁡x→0+s(x)=−1\displaystyle \lim_{x\to 0^+} s(x)=-1x→0+lim​s(x)=−1
  4. lim⁡x→0s(x)=DNE\displaystyle \lim_{x\to 0} s(x)=\text{DNE}x→0lim​s(x)=DNE (correct answer)
  5. lim⁡s(x)→0x=1\displaystyle \lim_{s(x)\to 0} x=1s(x)→0lim​x=1

Explanation: If left and right limits differ, the two-sided limit does not exist, denoted as DNE in notation. For s(x) = |x|/x, it approaches 1 from the right and -1 from the left as x nears 0, so \lim_{x \to 0} s(x) = DNE is accurate. This is valid when sides disagree, despite s(0) = 0. A common error is claiming a limit value like 0, confusing with the function at 0. Another symbolic mistake is reversing, such as \lim_{s(x) \to 0} x = 1, or using one-sided without specifying DNE for two-sided. Always check both directions for existence. Transferable notation checklist: 1. Write as \lim_{x \to a} f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.

Question 6

From the graph, H(x)H(x)H(x) approaches −2-2−2 as xxx approaches −1-1−1 from the right only. Which limit notation represents this?

  1. lim⁡x→−1+H(x)=−2\displaystyle \lim_{x\to -1^+} H(x)=-2x→−1+lim​H(x)=−2 (correct answer)
  2. lim⁡x→−1H(x)=−2\displaystyle \lim_{x\to -1} H(x)=-2x→−1lim​H(x)=−2
  3. H(−1)=−2\displaystyle H(-1)=-2H(−1)=−2
  4. lim⁡x→−1−H(x)=−2\displaystyle \lim_{x\to -1^-} H(x)=-2x→−1−lim​H(x)=−2
  5. lim⁡x→−2H(x)=−1\displaystyle \lim_{x\to -2} H(x)=-1x→−2lim​H(x)=−1

Explanation: Graph H(x) to -2 from right at -1. \lim_{x\to -1^+} H(x)=-2 correct. Valid right. Error: \lim_{x\to -1} H(x)=-2. H(-1)=-2 value. Specify. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 7

A graph indicates a(x)a(x)a(x) approaches −4-4−4 as xxx approaches 222 from both sides. Which limit notation matches this behavior?

  1. a(2)=−4\displaystyle a(2)=-4a(2)=−4
  2. lim⁡x→2a(x)=−4\displaystyle \lim_{x\to 2} a(x)=-4x→2lim​a(x)=−4 (correct answer)
  3. lim⁡x→2+a(x)=4\displaystyle \lim_{x\to 2^+} a(x)=4x→2+lim​a(x)=4
  4. lim⁡x→−4a(x)=2\displaystyle \lim_{x\to -4} a(x)=2x→−4lim​a(x)=2
  5. lim⁡x→2a(2)=−4\displaystyle \lim_{x\to 2} a(2)=-4x→2lim​a(2)=−4

Explanation: Limits capture behavior near points, graph showing a(x) to -4 at x=2 both sides. \lim_{x\to 2} a(x)=-4 matches. Valid ignoring a(2). Error: \lim_{x\to 2} a(2)=-4. a(2)=-4 is value. Distinguish. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 8

From the graph, v(x)v(x)v(x) approaches 000 as xxx approaches 111 from the left, while the right-hand behavior differs. Which expression matches?

  1. lim⁡x→1v(x)=0\displaystyle \lim_{x\to 1} v(x)=0x→1lim​v(x)=0
  2. v(1)=0\displaystyle v(1)=0v(1)=0
  3. lim⁡x→1−v(x)=0\displaystyle \lim_{x\to 1^-} v(x)=0x→1−lim​v(x)=0 (correct answer)
  4. lim⁡x→1+v(x)=0\displaystyle \lim_{x\to 1^+} v(x)=0x→1+lim​v(x)=0
  5. lim⁡x→0v(x)=1\displaystyle \lim_{x\to 0} v(x)=1x→0lim​v(x)=1

Explanation: One-sided notation is crucial when sides differ, as v(x) approaches 0 from the left at x=1, with differing right. Thus, \lim_{x\to 1^-} v(x)=0 is correct. This is valid for left-hand only. Error: \lim_{x\to 1} v(x)=0, implying both sides. v(1)=0 is value, not limit. Specify side. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 9

In the table, t(x)t(x)t(x) approaches 666 as xxx approaches 111. Which limit expression correctly represents this behavior?

  1. lim⁡x→1t(x)=6\displaystyle \lim_{x\to 1} t(x)=6x→1lim​t(x)=6 (correct answer)
  2. t(1)=6\displaystyle t(1)=6t(1)=6
  3. lim⁡x→6t(x)=1\displaystyle \lim_{x\to 6} t(x)=1x→6lim​t(x)=1
  4. lim⁡x→1−t(x)=5\displaystyle \lim_{x\to 1^-} t(x)=5x→1−lim​t(x)=5
  5. lim⁡x→1t(1)=6\displaystyle \lim_{x\to 1} t(1)=6x→1lim​t(1)=6

Explanation: Limit notation conveys approaching values, as the table shows t(x) nearing 6 as x approaches 1. Hence, lim⁡x→1t(x)=6\lim_{x\to 1} t(x)=6limx→1​t(x)=6 is appropriate for this behavior. This is valid regardless of t(1)t(1)t(1). An error is lim⁡x→1t(1)=6\lim_{x\to 1} t(1)=6limx→1​t(1)=6, misplacing evaluation. t(1)=6t(1)=6t(1)=6 is function value, not limit. Distinguish clearly. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x→ax \to ax→a. 3. Add +^{+}+ or −^{-}− for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 10

For x<0x<0x<0, t(x)=x2t(x)=x^2t(x)=x2 and for x>0x>0x>0, t(x)=3t(x)=3t(x)=3; which limit statement correctly represents the right-hand behavior as x→0x\to0x→0?

  1. lim⁡x→0t(x)=0\displaystyle \lim_{x\to0} t(x)=0x→0lim​t(x)=0
  2. t(0)=3\displaystyle t(0)=3t(0)=3
  3. lim⁡x→0+t(x)=3\displaystyle \lim_{x\to0^+} t(x)=3x→0+lim​t(x)=3 (correct answer)
  4. lim⁡x→0−t(x)=3\displaystyle \lim_{x\to0^-} t(x)=3x→0−lim​t(x)=3
  5. lim⁡x→0+t(x)=0\displaystyle \lim_{x\to0^+} t(x)=0x→0+lim​t(x)=0

Explanation: This piecewise function has t(x) = x² for x < 0 and t(x) = 3 for x > 0. The question asks specifically about right-hand behavior as x → 0. From the right (x > 0), we use t(x) = 3, so lim_{x→0^+} t(x) = 3. From the left (x < 0), we use t(x) = x², so lim_{x→0^-} t(x) = 0² = 0. The correct notation for the right-hand limit is lim_{x→0^+} t(x) = 3, properly indicating approach from the positive side. Option E incorrectly states the right-hand limit is 0, which would be the left-hand limit. A common error is confusing which formula applies for each direction. Limit notation checklist: x → 0^+ means x > 0 (approaching from the right), x → 0^- means x < 0 (approaching from the left), and match the correct piece of the function to each direction.

Question 11

The graph indicates n(x)n(x)n(x) approaches 555 as xxx approaches 222 from the right, while left-hand values approach 111. Which expression matches?

  1. lim⁡x→2n(x)=5\displaystyle \lim_{x\to 2} n(x)=5x→2lim​n(x)=5
  2. lim⁡x→2+n(x)=5\displaystyle \lim_{x\to 2^+} n(x)=5x→2+lim​n(x)=5 (correct answer)
  3. lim⁡x→2−n(x)=5\displaystyle \lim_{x\to 2^-} n(x)=5x→2−lim​n(x)=5
  4. n(2)=5\displaystyle n(2)=5n(2)=5
  5. lim⁡x→5n(x)=2\displaystyle \lim_{x\to 5} n(x)=2x→5lim​n(x)=2

Explanation: Graph n(x) to 5 from right at 2, left to 1. \lim_{x\to 2^+} n(x)=5 correct. Valid right. Error: \lim_{x\to 2} n(x)=5. n(2)=5 value. Specify. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 12

A table indicates h(x)h(x)h(x) approaches −2-2−2 as xxx approaches 000 from the right; which notation matches?

  1. lim⁡x→0+h(x)=−2\displaystyle \lim_{x\to 0^+} h(x)=-2x→0+lim​h(x)=−2 (correct answer)
  2. lim⁡x→0−h(x)=−2\displaystyle \lim_{x\to 0^-} h(x)=-2x→0−lim​h(x)=−2
  3. lim⁡x→0h(x)=2\displaystyle \lim_{x\to 0} h(x)=2x→0lim​h(x)=2
  4. lim⁡x→−2h(x)=0\displaystyle \lim_{x\to -2} h(x)=0x→−2lim​h(x)=0
  5. lim⁡x→0h(x)=h(0)\displaystyle \lim_{x\to 0} h(x)=h(0)x→0lim​h(x)=h(0)

Explanation: One-sided limit notation specifies direction using + for right or - for left, which is crucial when behavior differs on each side. The table shows h(x) approaching -2 only from the right as x nears 0, so \lim_{x \to 0^+} h(x) = -2 accurately represents this. This notation is valid as it matches the directional approach described. A common symbolic error is omitting the direction, like \lim_{x \to 0} h(x) = -2, assuming two-sided without evidence. Another mistake is confusing the limit with function equality, such as \lim_{x \to 0} h(x) = h(0). Always check if the data specifies one side or both. Transferable notation checklist: 1. Write as \lim_{x \to a} f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.

Question 13

For x≠5x\ne5x=5, u(x)=(x−5)(x+1)x−5u(x)=\dfrac{(x-5)(x+1)}{x-5}u(x)=x−5(x−5)(x+1)​ and u(5)=−3u(5)=-3u(5)=−3. Which expression represents u(x)u(x)u(x) as xxx approaches 555?

  1. lim⁡x→5u(x)=−3\lim_{x\to5} u(x)=-3limx→5​u(x)=−3
  2. lim⁡x→5u(x)=6\lim_{x\to5} u(x)=6limx→5​u(x)=6 (correct answer)
  3. lim⁡x→5−u(x)=5\lim_{x\to5^-} u(x)=5limx→5−​u(x)=5
  4. lim⁡x→5+u(x)=−6\lim_{x\to5^+} u(x)=-6limx→5+​u(x)=−6
  5. lim⁡x→5u(5)=−3\lim_{x\to5} u(5)=-3limx→5​u(5)=−3

Explanation: The function u(x)=(x−5)(x+1)x−5u(x) = \frac{(x-5)(x+1)}{x-5}u(x)=x−5(x−5)(x+1)​ simplifies to u(x)=x+1u(x) = x+1u(x)=x+1 for all x≠5x \neq 5x=5 by canceling the common factor (x−5)(x-5)(x−5). As xxx approaches 5, the simplified function approaches 5+1=65+1=65+1=6, regardless of the assigned value u(5)=−3u(5)=-3u(5)=−3. The correct limit notation is lim⁡x→5u(x)=6\lim_{x\to 5} u(x) = 6limx→5​u(x)=6. This illustrates a key principle: removable discontinuities don't affect limits—the limit exists even though the original expression is undefined at x=5x=5x=5. A common error is thinking the limit must equal the assigned function value, or writing lim⁡x→5u(5)\lim_{x\to 5} u(5)limx→5​u(5), which incorrectly evaluates the function inside the limit. Notation checklist: (1) lim⁡\limlim symbol, (2) x→5x\to 5x→5, (3) u(x)u(x)u(x) without substitution, (4) equals 6.

Question 14

For x≠0x\ne0x=0, s(x)=∣x∣xs(x)=\dfrac{|x|}{x}s(x)=x∣x∣​ and s(0)=0s(0)=0s(0)=0. Which expression represents the right-hand limit as xxx approaches 000?

  1. lim⁡x→0s(x)=0\lim_{x\to0} s(x)=0limx→0​s(x)=0
  2. lim⁡x→0+s(x)=1\lim_{x\to0^+} s(x)=1limx→0+​s(x)=1 (correct answer)
  3. lim⁡x→0−s(x)=1\lim_{x\to0^-} s(x)=1limx→0−​s(x)=1
  4. lim⁡x→0+s(x)=−1\lim_{x\to0^+} s(x)=-1limx→0+​s(x)=−1
  5. lim⁡x→0s(0)=0\lim_{x\to0} s(0)=0limx→0​s(0)=0

Explanation: The function s(x)=∣x∣xs(x) = \frac{|x|}{x}s(x)=x∣x∣​ equals xx=1\frac{x}{x} = 1xx​=1 when x>0x>0x>0 and −xx=−1\frac{-x}{x} = -1x−x​=−1 when x<0x<0x<0. For the right-hand limit as x→0+x\to 0^+x→0+, we consider positive values of xxx approaching 0, where s(x)=1s(x) = 1s(x)=1. Therefore, lim⁡x→0+s(x)=1\lim_{x\to 0^+} s(x) = 1limx→0+​s(x)=1 is the correct notation. The assigned value s(0)=0s(0)=0s(0)=0 is irrelevant to the limit calculation—limits describe behavior near a point, not at it. A common error is writing lim⁡x→0+s(0)\lim_{x\to 0^+} s(0)limx→0+​s(0), which incorrectly substitutes 0 into the function within the limit notation. Note that the left-hand limit would be -1, so the two-sided limit doesn't exist. Notation checklist: (1) lim⁡\limlim symbol, (2) x→0+x\to 0^+x→0+ for right approach, (3) s(x)s(x)s(x) not s(0)s(0)s(0), (4) equals 1.

Question 15

A table gives values of q(x)q(x)q(x) near x=3x=3x=3: q(2.9)=5.98q(2.9)=5.98q(2.9)=5.98, q(2.99)=5.998q(2.99)=5.998q(2.99)=5.998, q(3.01)=6.002q(3.01)=6.002q(3.01)=6.002, q(3.1)=6.02q(3.1)=6.02q(3.1)=6.02; which limit statement matches?

  1. q(3)=6\displaystyle q(3)=6q(3)=6
  2. lim⁡x→3q(x)=6\displaystyle \lim_{x\to3} q(x)=6x→3lim​q(x)=6 (correct answer)
  3. lim⁡x→6q(x)=3\displaystyle \lim_{x\to6} q(x)=3x→6lim​q(x)=3
  4. lim⁡x→3−q(x)=5.98\displaystyle \lim_{x\to3^-} q(x)=5.98x→3−lim​q(x)=5.98
  5. lim⁡x→3+q(x)=6.02\displaystyle \lim_{x\to3^+} q(x)=6.02x→3+lim​q(x)=6.02

Explanation: The table shows q(x) values approaching 6 as x approaches 3 from both sides: from the left (2.9 → 5.98, 2.99 → 5.998) and from the right (3.01 → 6.002, 3.1 → 6.02). Since the values approach 6 from both directions, the two-sided limit exists and equals 6. The correct notation is lim_{x→3} q(x) = 6, using the standard two-sided limit notation without directional superscripts. Option A incorrectly states q(3) = 6, which is a function value, not a limit statement. A common error is using overly specific decimal values (like 5.98 or 6.02) instead of recognizing the pattern approaching 6. Limit notation checklist: use lim_{x→a} for two-sided limits, omit superscripts when approaching from both sides, and identify the limiting value from the pattern in the table.

Question 16

For x≠4x\ne4x=4, r(x)=x−2x−4r(x)=\dfrac{\sqrt{x}-2}{x-4}r(x)=x−4x​−2​; which limit notation represents the value approached as x→4x\to4x→4?

  1. lim⁡x→4r(x)=14\displaystyle \lim_{x\to4} r(x)=\frac14x→4lim​r(x)=41​ (correct answer)
  2. r(4)=14\displaystyle r(4)=\frac14r(4)=41​
  3. lim⁡x→2r(x)=14\displaystyle \lim_{x\to2} r(x)=\frac14x→2lim​r(x)=41​
  4. lim⁡x→4−r(x)=−14\displaystyle \lim_{x\to4^-} r(x)=-\frac14x→4−lim​r(x)=−41​
  5. lim⁡x=4r(x)=14\displaystyle \lim_{x=4} r(x)=\frac14x=4lim​r(x)=41​

Explanation: The function r(x) = (√x - 2)/(x - 4) is undefined at x = 4, but we can find the limit by rationalizing. Multiplying by (√x + 2)/(√x + 2), we get r(x) = (x - 4)/[(x - 4)(√x + 2)] = 1/(√x + 2) for x ≠ 4. As x approaches 4, r(x) approaches 1/(√4 + 2) = 1/(2 + 2) = 1/4. The correct notation is lim_{x→4} r(x) = 1/4, using the two-sided limit since the simplified form approaches the same value from both directions. Option E incorrectly uses x = 4 instead of x → 4, which is improper limit notation. A common error is forgetting to simplify before evaluating the limit. Limit notation checklist: use → not = in limits, simplify indeterminate forms before evaluating, and use two-sided notation when both one-sided limits agree.

Question 17

For x<1x<1x<1, p(x)=1x−1p(x)=\dfrac{1}{x-1}p(x)=x−11​ and for x>1x>1x>1, p(x)=11−xp(x)=\dfrac{1}{1-x}p(x)=1−x1​; which limit notation describes p(x)p(x)p(x) as x→1−x\to1^-x→1−?

  1. lim⁡x→1−p(x)=−∞\displaystyle \lim_{x\to1^-} p(x)=-\inftyx→1−lim​p(x)=−∞ (correct answer)
  2. lim⁡x→1p(x)=0\displaystyle \lim_{x\to1} p(x)=0x→1lim​p(x)=0
  3. lim⁡x→1+p(x)=−∞\displaystyle \lim_{x\to1^+} p(x)=-\inftyx→1+lim​p(x)=−∞
  4. p(1)=−∞\displaystyle p(1)=-\inftyp(1)=−∞
  5. lim⁡x→1−p(x)=+∞\displaystyle \lim_{x\to1^-} p(x)=+\inftyx→1−lim​p(x)=+∞

Explanation: For this piecewise function, when x < 1, p(x) = 1/(x - 1), and as x approaches 1 from the left, the denominator (x - 1) approaches 0 through negative values. Since we're dividing 1 by increasingly small negative numbers, p(x) approaches -∞. The correct notation is lim_{x→1^-} p(x) = -∞, which properly indicates both the direction of approach (from the left) and the infinite behavior. Note that p(1) is undefined, so option D is incorrect notation. A common error is confusing the signs: as x → 1^-, we have x - 1 < 0, making 1/(x - 1) negative. Limit notation checklist: use lim notation for limits (not function notation), include direction superscripts for one-sided limits, and carefully track signs when dealing with infinite limits.

Question 18

The table indicates q(x)q(x)q(x) approaches 000 as xxx approaches −2-2−2. Which limit expression matches this behavior?

  1. lim⁡x→−2q(x)=0\displaystyle \lim_{x\to -2} q(x)=0x→−2lim​q(x)=0 (correct answer)
  2. q(−2)=0\displaystyle q(-2)=0q(−2)=0
  3. lim⁡x→0q(x)=−2\displaystyle \lim_{x\to 0} q(x)=-2x→0lim​q(x)=−2
  4. lim⁡x→−2+q(x)=2\displaystyle \lim_{x\to -2^+} q(x)=2x→−2+lim​q(x)=2
  5. lim⁡x→−2q(−2)=0\displaystyle \lim_{x\to -2} q(-2)=0x→−2lim​q(−2)=0

Explanation: Limit notation describes functional behavior near a point, with the table indicating q(x) approaches 0 as x nears -2. Thus, \lim_{x\to -2} q(x)=0 represents this two-sided approach. This is valid as it ignores q(-2) and focuses on vicinity. A common error is \lim_{x\to -2} q(-2)=0, blending notation improperly. q(-2)=0 denotes value at -2, not limit. Limits exist independently of point values. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 19

The table suggests c(x)c(x)c(x) approaches −1-1−1 as xxx approaches 444. Which limit expression correctly represents this behavior?

  1. c(4)=−1\displaystyle c(4)=-1c(4)=−1
  2. lim⁡x→4c(x)=−1\displaystyle \lim_{x\to 4} c(x)=-1x→4lim​c(x)=−1 (correct answer)
  3. lim⁡x→−1c(x)=4\displaystyle \lim_{x\to -1} c(x)=4x→−1lim​c(x)=4
  4. lim⁡x→4−c(x)=1\displaystyle \lim_{x\to 4^-} c(x)=1x→4−lim​c(x)=1
  5. lim⁡x→4c(4)=−1\displaystyle \lim_{x\to 4} c(4)=-1x→4lim​c(4)=−1

Explanation: Table suggests c(x) to -1 at x=4. \lim_{x\to 4} c(x)=-1 represents. Valid both sides. Error: \lim_{x\to 4} c(4)=-1. c(4)=-1 value. Separate. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 20

For g(x)=x2−9x−3g(x)=\frac{x^2-9}{x-3}g(x)=x−3x2−9​ when x≠3x\ne3x=3 and g(3)=10g(3)=10g(3)=10, which expression represents the limit as x→3x\to3x→3?

  1. lim⁡x→3g(x)=10\displaystyle \lim_{x\to 3} g(x)=10x→3lim​g(x)=10
  2. lim⁡x→10g(x)=3\displaystyle \lim_{x\to 10} g(x)=3x→10lim​g(x)=3
  3. lim⁡x→3g(x)=6\displaystyle \lim_{x\to 3} g(x)=6x→3lim​g(x)=6 (correct answer)
  4. lim⁡x→3−g(x)=10\displaystyle \lim_{x\to 3^-} g(x)=10x→3−lim​g(x)=10
  5. lim⁡g(x)→3x=6\displaystyle \lim_{g(x)\to 3} x=6g(x)→3lim​x=6

Explanation: Limit notation for rational functions often involves simplifying expressions to find the approaching value, ignoring the function's defined value at the point. For g(x) = (x^2 - 9)/(x - 3) simplified to x + 3 for x ≠ 3, the limit as x approaches 3 is 6, despite g(3) = 10, so \lim_{x \to 3} g(x) = 6 is correct. This is valid because limits consider behavior near the point, not at it. A common error is using the redefined value, like \lim_{x \to 3} g(x) = 10. Another mistake is reversing the limit, such as \lim_{x \to 10} g(x) = 3 or \lim_{g(x) \to 3} x = 6. The two-sided limit applies here as the function approaches the same value from both sides. Transferable notation checklist: 1. Write as \lim_{x \to a} f(x) = L, with x approaching a and f(x) to L. 2. Use + or - for one-sided limits if specified. 3. Do not equate limit to f(a) unless continuous. 4. Avoid swapping x and f(x) roles. 5. Confirm left and right agreement for two-sided limits.