Defining Limits and Using Limit Notation - AP Calculus AB

1. The graph indicates $n(x)$ approaches $5$ as $x$ approaches $2$ from the right, while left-hand values approach $1$. Which expression matches?
   Answer: $\displaystyle \lim_{x\to 2^+} n(x)=5$ 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.
2. From the graph, $p(x)$ approaches $-2$ as $x$ approaches $3$ from the left. Which limit notation represents this?
   Answer: $\displaystyle \lim_{x\to 3^-} p(x)=-2$ 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.
3. A function $s$ satisfies $s(x)=|x-5|$ for $x\ne5$ and $s(5)=2$; which limit expression matches the behavior as $x\to5$?
   Answer: $\displaystyle \lim_{x\to5} s(x)=0$ The function s(x) = |x - 5| for x ≠ 5 approaches 0 as x approaches 5 from either direction, since |5 - 5| = 0. Even though s(5) is defined as 2, this doesn't affect the limit behavior. The limit depends only on values near x = 5, not at x = 5 itself. The correct notation is $\lim_{x→5}$ s(x) = 0, which represents the value the function approaches. Option C incorrectly states the limit equals 2, confusing the limit with the function value at the point. A common error is thinking that if f(a) exists, then $\lim_{x→a}$ f(x) must equal f(a), but limits depend on nearby behavior, not the value at the point. Limit notation checklist: limits describe approaching behavior, function values describe actual values, and these can differ even when both exist.
4. A table indicates $h(x)$ approaches $-2$ as $x$ approaches $0$ from the right; which notation matches?
   Answer: $\displaystyle \lim_{x\to 0^+} h(x)=-2$ 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.
5. A function $r$ satisfies $r(x)=2$ for $x\le4$ and $r(x)=2+(x-4)$ for $x>4$. Which limit expression matches $r(x)$ as $x\to4^+$?
   Answer: $\lim_{x\to4^+} r(x)=2$ The function $r$ is defined as $r(x)=2$ for $x\leq 4$ and $r(x)=2+(x-4)$ for $x>4$. For the right-hand limit as $x\to 4^+$, we use the formula for $x>4$: as $x$ approaches 4 from the right, $(x-4)$ approaches 0, so $r(x)$ approaches $2+0=2$. The correct notation is $\lim_{x\to 4^+} r(x) = 2$, where the superscript plus indicates we only consider values greater than 4. Notice that both one-sided limits equal 2, making the function continuous at $x=4$. A common error is writing $\lim_{x\to 4^+} r(4)$, which incorrectly evaluates the function at 4 rather than describing the limiting behavior. Notation checklist: (1) $\lim$ symbol, (2) $x\to 4^+$ for right approach, (3) $r(x)$ not $r(4)$, (4) equals 2.
6. A table gives values of $q(x)$ near $x=3$: $q(2.9)=5.98$, $q(2.99)=5.998$, $q(3.01)=6.002$, $q(3.1)=6.02$; which limit statement matches?
   Answer: $\displaystyle \lim_{x\to3} q(x)=6$ 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.
7. From the graph, $H(x)$ approaches $-2$ as $x$ approaches $-1$ from the right only. Which limit notation represents this?
   Answer: $\displaystyle \lim_{x\to -1^+} H(x)=-2$ 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.
8. For $x\ne5$, $u(x)=\dfrac{(x-5)(x+1)}{x-5}$ and $u(5)=-3$. Which expression represents $u(x)$ as $x$ approaches $5$?
   Answer: $\lim_{x\to5} u(x)=6$ The function $u(x) = \frac{(x-5)(x+1)}{x-5}$ simplifies to $u(x) = x+1$ for all $x \neq 5$ by canceling the common factor $(x-5)$. As $x$ approaches 5, the simplified function approaches $5+1=6$, regardless of the assigned value $u(5)=-3$. The correct limit notation is $\lim_{x\to 5} u(x) = 6$. This illustrates a key principle: removable discontinuities don't affect limits—the limit exists even though the original expression is undefined at $x=5$. A common error is thinking the limit must equal the assigned function value, or writing $\lim_{x\to 5} u(5)$, which incorrectly evaluates the function inside the limit. Notation checklist: (1) $\lim$ symbol, (2) $x\to 5$, (3) $u(x)$ without substitution, (4) equals 6.
9. For $x\ne0$, $s(x)=\dfrac{|x|}{x}$ and $s(0)=0$. Which expression represents the right-hand limit as $x$ approaches $0$?
   Answer: $\lim_{x\to0^+} s(x)=1$ The function $s(x) = \frac{|x|}{x}$ equals $\frac{x}{x} = 1$ when $x>0$ and $\frac{-x}{x} = -1$ when $x<0$. For the right-hand limit as $x\to 0^+$, we consider positive values of $x$ approaching 0, where $s(x) = 1$. Therefore, $\lim_{x\to 0^+} s(x) = 1$ is the correct notation. The assigned value $s(0)=0$ is irrelevant to the limit calculation—limits describe behavior near a point, not at it. A common error is writing $\lim_{x\to 0^+} s(0)$, 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) $\lim$ symbol, (2) $x\to 0^+$ for right approach, (3) $s(x)$ not $s(0)$, (4) equals 1.
10. For $x\ne-1$, $q(x)=\dfrac{x^2-1}{x+1}$ and $q(-1)=10$. Which expression correctly represents $q(x)$ as $x$ approaches $-1$?
    Answer: $\lim_{x\to-1} q(x)=-2$ The function $q(x) = \frac{x^2-1}{x+1}$ can be simplified by factoring the numerator as $(x-1)(x+1)$, giving us $q(x) = \frac{(x-1)(x+1)}{x+1} = x-1$ for all $x \neq -1$. As $x$ approaches $-1$, the simplified function approaches $-1-1 = -2$, regardless of the assigned value $q(-1)=10$. The correct limit notation is $\lim_{x\to -1} q(x) = -2$. A common error is thinking the limit must equal the function value at the point, but limits describe nearby behavior, not the value at the point itself. Another mistake is writing $\lim_{x\to -1} q(-1)$, which incorrectly evaluates the function inside the limit. Notation checklist: (1) $\lim$ symbol, (2) $x\to -1$, (3) $q(x)$ without substitution, (4) equals -2.
11. The table indicates $q(x)$ approaches $0$ as $x$ approaches $-2$. Which limit expression matches this behavior?
    Answer: $\displaystyle \lim_{x\to -2} q(x)=0$ 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.
12. For $x<0$, $h(x)=2x+1$; for $x\ge0$, $h(x)=x^2+1$. Which expression represents the left-hand limit as $x$ approaches $0$?
    Answer: $\lim_{x\to0^-} h(x)=1$ For the piecewise function $h$, we need the left-hand limit as $x$ approaches 0, meaning we consider values where $x<0$. In this region, $h(x) = 2x+1$, so as $x$ approaches 0 from the left, $h(x)$ approaches $2(0)+1 = 1$. The correct notation is $\lim_{x\to 0^-} h(x) = 1$, where the superscript minus indicates left-hand approach. A common error is writing $\lim_{x\to 0^-} h(0)$, which incorrectly places the function value inside the limit notation—limits describe the function's behavior, not a single value. Note that the right-hand limit would also equal 1 since $h(0) = 0^2+1 = 1$, making the two-sided limit exist. Notation checklist: (1) $\lim$ symbol, (2) $x\to 0^-$ for left approach, (3) $h(x)$ not $h(0)$, (4) equals 1.
13. For $x\ne2$, $f(x)=\dfrac{x^2-4}{x-2}$ and $f(2)=7$. Which expression represents $f(x)$ as $x$ approaches $2$?
    Answer: $\lim_{x\to2} f(x)=4$ The function $f(x) = \frac{x^2-4}{x-2}$ can be simplified by factoring the numerator as $(x-2)(x+2)$, giving us $f(x) = \frac{(x-2)(x+2)}{x-2} = x+2$ for all $x \neq 2$. As $x$ approaches 2, the function approaches $2+2=4$, regardless of what value is assigned at $x=2$. The limit notation $\lim_{x\to 2} f(x) = 4$ correctly captures this behavior. A common error is confusing the limit with the function value at a point—notice that $f(2)=7$ but this doesn't affect the limit. Remember: limits describe behavior near a point, not at the point itself. Notation checklist: (1) $\lim$ symbol present, (2) approach variable and value specified, (3) function without substitution, (4) equals the limiting value.
14. For $x<1$, $p(x)=\dfrac{1}{x-1}$ and for $x>1$, $p(x)=\dfrac{1}{1-x}$; which limit notation describes $p(x)$ as $x\to1^-$?
    Answer: $\displaystyle \lim_{x\to1^-} p(x)=-\infty$ 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.
15. For $x\ne2$, $v(x)=\dfrac{x^2-4}{x-2}$ and $v(2)=10$; which limit notation matches the value approached as $x\to2$?
    Answer: $\displaystyle \lim_{x\to2} v(x)=4$ The function v(x) = (x² - 4)/(x - 2) is undefined at x = 2, but we can find the limit by factoring. Since x² - 4 = (x + 2)(x - 2), we get v(x) = (x + 2)(x - 2)/(x - 2) = x + 2 for x ≠ 2. As x approaches 2, v(x) approaches 2 + 2 = 4. The fact that v(2) = 10 doesn't affect the limit, which depends only on nearby values. The correct notation is $\lim_{x→2}$ v(x) = 4, representing the value approached regardless of v(2). Option B incorrectly states the limit is 10, confusing the defined function value with the limit. A common error is thinking limits must equal function values when both exist. Limit notation checklist: limits describe approaching behavior, simplified forms reveal limit values, and limits can differ from function values at a point.
16. For $x\ne-1$, $g(x)=\dfrac{x^2-1}{x+1}$; which expression represents the value $g(x)$ approaches as $x\to-1$?
    Answer: $\displaystyle \lim_{x\to-1} g(x)=-2$ The function g(x) = (x² - 1)/(x + 1) is undefined at x = -1, but we can find the limit by factoring. Since x² - 1 = (x + 1)(x - 1), we get g(x) = (x + 1)(x - 1)/(x + 1) = x - 1 for x ≠ -1. As x approaches -1, g(x) approaches (-1) - 1 = -2. The correct notation is $\lim_{x→-1}$ g(x) = -2, using the two-sided limit since the simplified function approaches the same value from both sides. Note that g(-1) is undefined, so option A is incorrect. A common error is confusing the limit value with the function value at a point. Limit notation checklist: use $\lim_{x→a}$ for limits, f(a) for function values, and remember limits can exist even when f(a) is undefined.
17. For $x<0$, $t(x)=x^2$ and for $x>0$, $t(x)=3$; which limit statement correctly represents the right-hand behavior as $x\to0$?
    Answer: $\displaystyle \lim_{x\to0^+} t(x)=3$ 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.
18. For $g(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $g(3)=10$, which expression represents the limit as $x\to3$?
    Answer: $\displaystyle \lim_{x\to 3} g(x)=6$ 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.
19. A function $f$ satisfies $f(x)=3x+1$ for $x<2$ and $f(x)=x^2-3$ for $x>2$; which limit notation matches $f$ as $x\to2$?
    Answer: $\displaystyle \lim_{x\to2^-} f(x)=7$ This piecewise function has different formulas on either side of x = 2, so we must check one-sided limits. For x < 2, we use f(x) = 3x + 1, giving $\lim_{x→2^-}$ f(x) = 3(2) + 1 = 7. For x > 2, we use f(x) = x² - 3, giving $\lim_{x→2^+}$ f(x) = (2)² - 3 = 1. Since the left and right limits differ (7 ≠ 1), the two-sided limit does not exist. The correct notation is $\lim_{x→2^-}$ f(x) = 7, which specifically indicates approaching from the left. A common error is writing $\lim_{x=2}$ (using equals instead of arrow), which is improper notation. Limit notation checklist: use → not =, specify direction (^- or ^+) for one-sided limits, and verify both sides exist for two-sided limits.
20. A graph indicates $a(x)$ approaches $-4$ as $x$ approaches $2$ from both sides. Which limit notation matches this behavior?
    Answer: $\displaystyle \lim_{x\to 2} a(x)=-4$ 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.

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