Algebraic Properties of Limits - AP Calculus BC

1. A projectile's height is $h(t)=\dfrac{t^2+3t}{t+3}$; compute $\lim_{t\to 1} h(t)$.
   Answer: $\dfrac{1^2+3(1)}{1+3}$ This problem requires using the limit laws for rational functions. Since the denominator at t=1 is 1+3=4 ≠0, direct substitution is valid. The limit is ($1^2$ +3*1)/(1+3) = (1+3)/4 =1. The choice B is this plugged-in expression. A tempting distractor is D dfrac{1^2 +3(1)}{3} =4/3, perhaps from misplacing the denominator as 3 alone. A transferable strategy is to evaluate the denominator's limit first to ensure it's not zero before applying direct substitution.
2. A temperature adjustment is $A(x)=7-\dfrac{2}{x+4}$; compute $\lim_{x\to 0} A(x)$.
   Answer: $7-\dfrac{2}{4}$ This problem requires using the limit laws for constants, quotients, and sums. The function is continuous at x=0, so direct substitution gives 7 - 2/(0+4) =7 -2/4 =7-0.5=6.5. The choice A is this expression. It follows from lim (constant) = constant, and lim 2/(x+4) =2/4. A tempting distractor is C 7 -2/0, perhaps from forgetting the +4 in denominator, leading to undefined. A transferable strategy is to substitute into each part of the expression using the appropriate limit laws.
3. If $m(x)=\dfrac{3x-6}{x+5}+2$, find $\lim_{x\to 1} m(x)$.
   Answer: $\dfrac{3(1)-6}{1+5}+2$ This problem requires using the limit laws for sums and quotients. The function is continuous at x=1 since denominator 1+5=6 ≠0. The limit is [3(1)-6]/(1+5) +2 = (-3)/6 +2 = -0.5 +2=1.5. The choice A is this expression. A tempting distractor is B dfrac{3(1-6)}{1+5}+2 = 3*(-5)/6 +2 = -15/6 +2 = -2.5+2=-0.5, from misplacing the parentheses in numerator. A transferable strategy is to apply substitution to each component and combine using the sum law.
4. A model uses $B(x)=\dfrac{1}{x^2+1}+\dfrac{x}{2}$; evaluate $\lim_{x\to 2} B(x)$.
   Answer: $\dfrac{1}{5}+1$ This problem requires applying the algebraic properties of limits, such as the sum and quotient rules. B(x) = 1/($x^2$ +1) + x/2 is continuous at x=2, so the limit is B(2). Substituting x=2 gives 1/(4 +1) + 2/2 = 1/5 + 1. This follows from quotient rules for each term and sum property. A tempting distractor like 1/5 + 1/2 might halve incorrectly, but it's 1 + 0.2 = 1.2. A transferable strategy is to compute each addend's limit separately.
5. A cost adjustment is $A(x)=(x-1)(x+5)$. Compute $\lim_{x\to -2} A(x)$ using algebraic properties.
   Answer: $-9$ This limit problem involves applying limit laws to a factored polynomial expression. Since polynomials are continuous everywhere, we can use direct substitution: $\lim_{x\to -2} (x-1)(x+5) = (-2-1)(-2+5) = (-3)(3) = -9$. The limit laws tell us we can evaluate the limit of a product by taking the product of the individual limits. A common mistake would be to get $9$ by losing track of the negative signs, computing $3 \times 3$ instead of $(-3) \times 3$. When evaluating limits of products, carefully track the sign of each factor, especially when substituting negative values.
6. A temperature model uses $T(t)=3t^2-5t+8$. Find $\lim_{t\to 2} T(t)$ using limit laws.
   Answer: $10$ This problem requires applying limit laws to evaluate the limit of a polynomial function. Since polynomials are continuous everywhere, we can use direct substitution: $\lim_{t\to 2} (3t^2-5t+8) = 3(2)^2-5(2)+8 = 3(4)-10+8 = 12-10+8 = 10$. The limit laws tell us we can evaluate the limit of each term separately and combine them. A common error would be to substitute $t=2$ into only part of the expression, perhaps getting $12-10 = 2$ by forgetting the constant term. When evaluating limits of polynomials, always substitute the value directly into the entire expression since polynomials are continuous everywhere.
7. A concentration is $k(x)=\dfrac{x^2-4x+1}{x}$; find $\lim_{x\to 5} k(x)$.
   Answer: $\dfrac{25-20+1}{5}$ This problem requires using the limit laws for rational functions. The denominator at x=5 is 5 ≠0, so direct substitution is applicable. The limit is ($5^2$ -4*5 +1)/5 = (25-20+1)/5 =6/5. The choice A is this expression. A tempting distractor is D dfrac{25-4(5)+1}{0}, perhaps from incorrectly using 0 in denominator. A transferable strategy is to substitute the value into both numerator and denominator separately and then divide.
8. If $f(x)=\dfrac{5x-1}{2x+3}$ models efficiency, determine $\lim_{x\to -1} f(x)$.
   Answer: $\dfrac{-5-1}{-2+3}$ This problem requires using the limit laws for quotients. The denominator at x=-1 is 2*(-1)+3=-2+3=1 ≠0, so direct substitution works. The limit is (5*(-1)-1)/(2*(-1)+3) = (-5-1)/(-2+3) = -6/1 = -6. The choice D represents this expression with the signs correctly applied. A tempting distractor is C dfrac{5-1}{2+3}=4/5, which ignores the negative in x=-1. A transferable strategy is to carefully plug in the value with attention to negative signs and order of operations.
9. Given $R(x)=\dfrac{(x+1)(x-3)}{x-3}$ for $x\ne 3$, evaluate $\lim_{x\to 3} R(x)$.
   Answer: $3+1$ This problem requires using the algebraic properties of limits, including simplifying rational expressions before taking the limit. The function R(x) = (x+1)(x-3)/(x-3) simplifies to x+1 for x≠3. Therefore, the limit as x approaches 3 is 3+1=4. The choice B is this value. A tempting distractor is A, which plugs into the original to get 0/0, an indeterminate form that doesn't give the limit. A transferable strategy is to factor and cancel common factors in rational functions when direct substitution gives 0/0.
10. For $g(x)=\dfrac{x^2+1}{x^2+4}$, determine $\lim_{x\to -2} g(x)$ using algebraic limit properties.
    Answer: $\dfrac{(-2)^2+1}{(-2)^2+4}$ This problem requires using the algebraic limit properties for quotients of polynomials. Since denominator at x=-2 is $(-2)^2$ +4=4+4=8 ≠0, direct substitution applies. The limit is [$(-2)^2$ +1]/[$(-2)^2$ +4] = (4+1)/(4+4)=5/8. The choice A uses the correct order of operations with $(-2)^2$=4. A tempting distractor is C dfrac{-4+1}{-4+4} = -3/0, from incorrectly computing -$2^2$ as -4 instead of (+4). A transferable strategy is to remember that $(-a)^2$ = $a^2$ is positive, and substitute carefully.
11. A signal is modeled by $f(x)=\dfrac{5-2x}{x^2+1}$. Find $\lim_{x\to 0} f(x)$.
    Answer: $5$ This problem requires using limit laws on a rational function where the denominator is non-zero at the limit point. Since $0^2+1 = 1 \neq 0$, we can apply direct substitution: $\lim_{x\to 0} \frac{5-2x}{x^2+1} = \frac{5-2(0)}{0^2+1} = \frac{5-0}{0+1} = \frac{5}{1} = 5$. The limit laws allow us to evaluate the numerator and denominator separately when the denominator limit is non-zero. A potential error would be to think the limit is 0 because $x$ approaches 0, forgetting that we need to evaluate the entire expression. For rational functions, always substitute the limit value into both numerator and denominator, checking that the denominator remains non-zero.
12. A growth function is $P(x)=7-\dfrac{3}{x}$. Determine $\lim_{x\to 6} P(x)$.
    Answer: $\dfrac{13}{2}$ This problem uses limit laws to evaluate a function involving subtraction of a rational term. Since $x=6$ makes the denominator non-zero, we can use direct substitution: $\lim_{x\to 6} \left(7-\frac{3}{x}\right) = 7-\frac{3}{6} = 7-\frac{1}{2} = \frac{14}{2}-\frac{1}{2} = \frac{13}{2}$. The limit laws tell us we can evaluate the limit of a difference by taking the difference of the individual limits. A potential mistake would be to compute $7-3 = 4$ and then divide by 6, getting $\frac{4}{6} = \frac{2}{3}$, by misapplying the order of operations. For expressions with multiple operations, always follow the correct order: division before subtraction.
13. A concentration is modeled by $c(x)=\dfrac{x^2+1}{x^2+4}$; compute $\lim_{x\to 2} c(x)$.
    Answer: $\dfrac{5}{8}$ This problem requires applying the algebraic properties of limits, including the quotient rule. $c(x) = \dfrac{x^2 + 1}{x^2 + 4}$ has denominator limit at x=2 of $8 \neq 0$, so direct substitution applies. Plugging in x=2 gives $(4 + 1)/(4 + 4) = \dfrac{5}{8}$. This uses sum rules in numerator and denominator before the quotient. A tempting distractor like $\dfrac{5}{4}$ might swap the constants, but correct addition yields $\dfrac{5}{8}$. A transferable strategy is to confirm the denominator's limit is nonzero for rational functions before substituting.
14. In a mixing process, $C(t)=\sqrt{t+9}-2$. Find $\lim_{t\to 7} C(t)$.
    Answer: $2$ This limit problem requires applying limit laws to a composite function involving a square root. Since the function $C(t) = \sqrt{t+9}-2$ is continuous at $t=7$ (the expression under the square root is positive), we can use direct substitution. We get $\lim_{t\to 7} C(t) = \sqrt{7+9}-2 = \sqrt{16}-2 = 4-2 = 2$. A tempting error would be to compute $\sqrt{7}-2$, which would give approximately $-0.35$, by forgetting to add 9 to the input first. When evaluating limits involving composite functions, work from the inside out and ensure each operation is valid at the limit point.
15. For a revenue function $R(x)=\dfrac{2x^2-3x+1}{x}$, evaluate $\lim_{x\to 4} R(x)$.
    Answer: $\dfrac{21}{4}$ This problem involves using limit laws to evaluate a rational function at a point where it's defined. First, we simplify the expression: $R(x) = \frac{2x^2-3x+1}{x} = 2x-3+\frac{1}{x}$ for $x \neq 0$. Since $x=4$ is in the domain, we can apply direct substitution: $\lim_{x\to 4} R(x) = 2(4)-3+\frac{1}{4} = 8-3+\frac{1}{4} = 5+\frac{1}{4} = \frac{21}{4}$. A common mistake would be to evaluate only the numerator at $x=4$ getting $21$, forgetting to divide by 4. For rational functions, always check that the denominator is non-zero at the limit point before using direct substitution.
16. A function is $m(x)=\dfrac{5}{2}x^2-4x+9$; evaluate $\lim_{x\to 0} m(x)$.
    Answer: $9$ This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules. m(x) is a polynomial, continuous everywhere, so the limit as x approaches 0 is m(0). Substituting x=0 gives (5/2)*0 - 4*0 + 9 = 9. This follows from applying the power rule to each term and combining with sum properties. A tempting distractor like 0 might come from only considering the variable terms, but the constant 9 remains. A transferable strategy is to evaluate polynomials at the limit point directly due to their continuity.
17. During a lab, $g(x)=\dfrac{2x^2+3x-5}{x-1}$ is recorded; evaluate $\lim_{x\to 1} (2x^2+3x-5)$.
    Answer: $0$ This problem illustrates the use of limit laws, specifically the sum, difference, and constant multiple rules for limits. The expression $2x^2$ + 3x - 5 is a polynomial, so the limit as x approaches 1 is simply its value at x=1. Substituting x=1 gives 2*$(1)^2$ + 3*1 - 5 = 2 + 3 - 5 = 0. This follows from applying the power rule to each term and then using the sum and difference properties of limits. A tempting distractor like -4 might come from confusing it with evaluating g(x), but the question asks only for the numerator, which is defined at x=1. A transferable strategy is to apply direct substitution for polynomials, as they are continuous everywhere.
18. A particle's position is $s(t)=\dfrac{t^2+1}{t+1}$. Determine $\lim_{t\to 3} s(t)$.
    Answer: $\dfrac{5}{2}$ This problem uses limit laws to evaluate a rational function where direct substitution is valid. Since the denominator $t+1$ is non-zero when $t=3$, we can substitute directly: $\lim_{t\to 3} \frac{t^2+1}{t+1} = \frac{3^2+1}{3+1} = \frac{9+1}{4} = \frac{10}{4} = \frac{5}{2}$. The limit laws allow us to evaluate the numerator and denominator limits separately when the denominator limit is non-zero. A common error would be to simplify incorrectly, perhaps getting $\frac{10}{3}$ by mistakenly using $t$ instead of $t+1$ in the denominator. For rational functions, always verify the denominator is non-zero at the limit point before applying direct substitution.
19. For $y(x)=\dfrac{2x+5}{x^2+9}$, compute $\lim_{x\to -2} \dfrac{2x+5}{x^2+9}$.
    Answer: $\dfrac{1}{13}$ This problem illustrates the use of limit laws, including the quotient rule. y(x) = (2x + 5)/($x^2$ + 9) has denominator at x=-2 of 13 ≠ 0, so direct substitution works. Plugging in x=-2 gives (2*(-2) + 5)/(4 + 9) = 1/13. This uses sum in numerator and denominator before quotient. A tempting distractor like -9/13 might misadd the numerator as -4 -5, but it's +5, yielding positive. A transferable strategy is to carefully compute arithmetic in numerators and denominators for rational limits.
20. For $g(h)=\dfrac{(h+2)^3}{4h-1}$ in a lab model, evaluate $\lim_{h\to 1} g(h)$.
    Answer: $9$ This limit involves applying limit laws to a rational function with a cubic numerator. First, we verify the denominator is non-zero: $4(1)-1 = 3 \neq 0$, so direct substitution is valid. Computing: $\lim_{h\to 1} \frac{(h+2)^3}{4h-1} = \frac{(1+2)^3}{4(1)-1} = \frac{3^3}{3} = \frac{27}{3} = 9$. The limit laws allow us to evaluate powers and quotients when all operations are defined. A common error would be to compute $(1+2)^3 = 9$ instead of $27$, confusing squaring with cubing. When evaluating limits involving powers, carefully compute each exponent and simplify the resulting fraction completely.

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