Algebraic Properties of Limits - AP Calculus AB

1. A temperature model is $T(t)=t^3-2t$. Evaluate $\lim_{t\to -1} T(t)$.
   Answer: 1 This cubic function is continuous everywhere, allowing direct substitution. Using limit laws: $\lim_{t\to -1} T(t) = (-1)^3 - 2(-1) = -1 + 2 = 1$. We apply the difference rule and power rule, carefully handling the negative input: $(-1)^3 = -1$ and $-2(-1) = +2$. A common mistake would be sign errors when working with negative values, particularly $(-1)^3 = -1$ or the double negative in $-2(-1)$. The transferable strategy is to work methodically with signs when substituting negative values into polynomial expressions.
2. Given $g(x)=2\sin(x)+1$, find $\lim_{x\to 0} g(x)$ using basic limit laws.
   Answer: 1 This trigonometric function combines basic trigonometric limits with algebraic operations. Using the sum rule and scalar multiplication rule: $\lim_{x\to 0} g(x) = \lim_{x\to 0} 2\sin(x) + \lim_{x\to 0} 1 = 2\lim_{x\to 0} \sin(x) + 1$. Since $\lim_{x\to 0} \sin(x) = 0$, we get $2(0) + 1 = 1$. A common error would be incorrectly recalling that $\lim_{x\to 0} \sin(x) = 1$ instead of 0, or confusing this with $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$. The transferable strategy is to memorize fundamental trigonometric limits and apply limit laws systematically.
3. Given $h(x)=\cos(x)-2\sin(x)$, evaluate $\lim_{x\to 0} h(x)$ using limit laws.
   Answer: 1 This function combines basic trigonometric functions with different coefficients. Using limit laws and the fact that $\sin(0) = 0$ and $\cos(0) = 1$: $\lim_{x\to 0} h(x) = \cos(0) - 2\sin(0) = 1 - 2(0) = 1 - 0 = 1$. We apply the difference rule and scalar multiplication rule systematically. A common mistake would be confusing the values of $\sin(0)$ and $\cos(0)$ or incorrectly applying the coefficient to the cosine term. The transferable strategy is to memorize fundamental trigonometric limits and apply limit laws term by term.
4. A function is $r(x)=7+\frac{x}{3}$. Find $\lim_{x\to 6} r(x)$ using limit laws.
   Answer: 9 This linear function with a fractional coefficient is continuous everywhere. Using direct substitution: $\lim_{x\to 6} r(x) = 7 + \frac{6}{3} = 7 + 2 = 9$. We apply the sum rule for limits, evaluating each term separately: the constant term gives 7, and the rational term gives $\frac{6}{3} = 2$. A common error would be arithmetic mistakes in the fraction division or not recognizing this as a continuous function. The key strategy is to identify continuous functions and apply direct substitution, treating each term according to the appropriate limit law.
5. Let f and g be functions where $\lim_{x \to 2^+} f(x) = 5$ and $\lim_{x \to 2^+} g(x) = -3$. What is the value of $\lim_{x \to 2^+} \frac{f(x) + 1}{g(x)}$?
   Answer: $-2$ The standard limit properties apply to one-sided limits. Thus, $\lim_{x \to 2^+} \frac{f(x) + 1}{g(x)} = \frac{\lim_{x \to 2^+} (f(x) + 1)}{\lim_{x \to 2^+} g(x)} = \frac{\lim_{x \to 2^+} f(x) + \lim_{x \to 2^+} 1}{\lim_{x \to 2^+} g(x)} = \frac{5+1}{-3} = \frac{6}{-3} = -2$. Distractor B subtracts 1 in the numerator instead of adding. Distractor C is a sign error. Distractor D is for students who do not realize properties apply to one-sided limits.
6. Given $u(x)=\sin(x)+\cos(x)$, evaluate $\lim_{x\to 0} u(x)$ using limit laws.
   Answer: 1 This function combines two basic trigonometric functions. Using the sum rule and the fundamental limits $\lim_{x\to 0} \sin(x) = 0$ and $\lim_{x\to 0} \cos(x) = 1$: $\lim_{x\to 0} u(x) = \lim_{x\to 0} \sin(x) + \lim_{x\to 0} \cos(x) = 0 + 1 = 1$. Both sine and cosine are continuous at $x = 0$, allowing direct evaluation. A common mistake would be confusing the values of $\sin(0)$ and $\cos(0)$ or incorrectly recalling these fundamental limits. The transferable strategy is to memorize key trigonometric values and apply the sum rule systematically for combinations of trigonometric functions.
7. Let f and g be functions such that $\lim_{x \to 3} f(x) = 5$ and $\lim_{x \to 3} g(x) = -2$. What is the value of $\lim_{x \to 3} [2f(x) - g(x)]$?
   Answer: $12$ Using the sum, difference, and constant multiple properties of limits: $\lim_{x \to 3} [2f(x) - g(x)] = 2 \lim_{x \to 3} f(x) - \lim_{x \to 3} g(x) = 2(5) - (-2) = 10 + 2 = 12$.
8. Given $\lim_{x \to 9} f(x) = 4$, what is the value of $\lim_{x \to 9} \frac{\sqrt{f(x)}}{x}$?
   Answer: $\frac{2}{9}$ Using the quotient and composite function properties: $\lim_{x \to 9} \frac{\sqrt{f(x)}}{x} = \frac{\lim_{x \to 9} \sqrt{f(x)}}{\lim_{x \to 9} x} = \frac{\sqrt{\lim_{x \to 9} f(x)}}{9} = \frac{\sqrt{4}}{9} = \frac{2}{9}$. Distractor B results from forgetting the square root in the numerator. Distractor C results from taking the square root of the denominator as well. Distractor D results from ignoring the denominator.
9. If $\lim_{x \to a} f(x) = -3$, what is the value of $\lim_{x \to a} [f(x)]^4$?
   Answer: $81$ By the power property of limits, $\lim_{x \to a} [f(x)]^4 = \left(\lim_{x \to a} f(x)\right)^4 = (-3)^4 = 81$. Distractor A makes a common sign error, forgetting that an even power results in a positive number. Distractors B and C result from confusing exponentiation with multiplication ($4 \cdot (-3) = -12$ and $4 \cdot 3 = 12$).
10. Let $f(x)=2^x+3$. Evaluate $\lim_{x\to 2} f(x)$ using limit laws.
    Answer: 7 This exponential function is continuous everywhere, allowing direct substitution. Using limit laws: $\lim_{x\to 2} f(x) = 2^2 + 3 = 4 + 3 = 7$. We apply the sum rule, with the exponential term $2^2 = 4$ and the constant term 3. A common mistake would be confusing exponential and polynomial notation or making arithmetic errors. The transferable strategy is to recognize that exponential functions are continuous everywhere, so direct substitution applies, and to compute exponential values carefully.
11. If $\lim_{x \to 1} f(x) = 8$, what is the value of $\lim_{x \to 1} (x^3 \sqrt{f(x)})$?
    Answer: $2\sqrt{2}$ Using the product and composite function properties of limits: $\lim_{x \to 1} (x^3 \sqrt{f(x)}) = (\lim_{x \to 1} x^3) \cdot (\lim_{x \to 1} \sqrt{f(x)}) = (1^3) \cdot \sqrt{\lim_{x \to 1} f(x)} = 1 \cdot \sqrt{8} = 2\sqrt{2}$. Distractor A incorrectly computes $\sqrt[3]{8}$ instead of $\sqrt{8}$. Distractor C results from forgetting the square root: $1 \cdot 8 = 8$. Distractor D results from multiplying the limit by itself ($8 \cdot \sqrt{1}$).
12. Let f and g be functions such that $\lim_{x \to 2} g(x) = 3$. If f is continuous at $x=3$ and $f(3) = 5$, what is $\lim_{x \to 2} f(g(x))$?
    Answer: $5$ Because f is continuous at $\lim_{x \to 2} g(x) = 3$, we can use the property for the limit of a composite function: $\lim_{x \to 2} f(g(x)) = f(\lim_{x \to 2} g(x)) = f(3) = 5$. Distractor B is the limit of the inner function, not the composite function. Distractor D is the sum of the input and output values ($3+5$).
13. Suppose $\lim_{x \to c} f(x) = L$, where L is a finite real number, and $\lim_{x \to c} g(x)$ does not exist. Which of the following limits must also not exist?
    Answer: $\lim_{x \to c} [f(x) + g(x)]$ If $\lim_{x \to c} [f(x) + g(x)]$ were to exist (call it M), then $\lim_{x \to c} g(x) = \lim_{x \to c} ([f(x) + g(x)] - f(x)) = M - L$, which would exist. This is a contradiction. Therefore, the limit of the sum must not exist. For A, the limit could exist if $L=0$. For B, the limit could exist if, for example, $f(x)$ was a large multiple of $g(x)$. For D, the limit could exist; for example, if $g(x)$ alternates between 1 and -1.
14. If $f(x)=\frac{2x-1}{x+1}$, evaluate $\lim_{x\to 1} f(x)$ using limit laws.
    Answer: $\frac{1}{2}$ This rational function is continuous at $x = 1$ since $1 + 1 = 2 \neq 0$. Using the quotient rule: $\lim_{x\to 1} f(x) = \frac{\lim_{x\to 1}(2x - 1)}{\lim_{x\to 1}(x + 1)} = \frac{2(1) - 1}{1 + 1} = \frac{2 - 1}{2} = \frac{1}{2}$. Both numerator and denominator limits exist with the denominator non-zero. A common error would be arithmetic mistakes in $2 - 1 = 1$ or $1 + 1 = 2$. The key strategy is to evaluate numerator and denominator separately, then form the fraction and simplify if necessary.
15. A function is $M(x)=\sqrt{25+4x}$. Evaluate $\lim_{x\to 0} M(x)$ using limit laws.
    Answer: 5 This square root function is continuous at $x = 0$ since $25 + 4(0) = 25 > 0$. Using direct substitution: $\lim_{x\to 0} M(x) = \sqrt{25 + 4(0)} = \sqrt{25 + 0} = \sqrt{25} = 5$. The expression under the radical is positive, ensuring the function is well-defined. A common mistake would be arithmetic errors in $4(0) = 0$ or $\sqrt{25} = 5$. The key strategy is to verify that expressions under radicals remain non-negative, then apply direct substitution with careful arithmetic.
16. If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, where L and M are finite real numbers, which of the following limit properties is NOT always true without an additional condition?
    Answer: $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ The quotient property for limits, stated in choice D, is only guaranteed to be true if the limit of the denominator is non-zero, i.e., $M \neq 0$. Since this condition is not stated, this property is not always true. The sum, product, and power properties stated in A, B, and C are always true when the individual limits exist and are finite.
17. A function is $w(x)=\frac{3x}{x+1}$. Evaluate $\lim_{x\to 2} w(x)$ using limit laws.
    Answer: 2 This rational function is continuous at $x = 2$ since $2 + 1 = 3 \neq 0$. Using the quotient rule: $\lim_{x\to 2} w(x) = \frac{\lim_{x\to 2}(3x)}{\lim_{x\to 2}(x + 1)} = \frac{3(2)}{2 + 1} = \frac{6}{3} = 2$. Both numerator and denominator limits exist, and the denominator is non-zero, allowing direct application of the quotient rule. A common mistake would be not checking that the denominator is non-zero or making arithmetic errors in the fraction simplification. The transferable approach is to verify continuity, then apply the quotient rule systematically.
18. Given $m(x)=\sqrt{16-x}$, find $\lim_{x\to 7} m(x)$ using limit laws.
    Answer: 3 This square root function is continuous at $x = 7$ since $16 - 7 = 9 > 0$. Using the composition rule for limits: $\lim_{x\to 7} m(x) = \sqrt{\lim_{x\to 7}(16 - x)} = \sqrt{16 - 7} = \sqrt{9} = 3$. The expression under the radical approaches a positive value, so the limit exists and equals the square root of that value. A common error would be arithmetic mistakes in $16 - 7$ or not checking that the expression under the radical is non-negative. The key strategy is to verify the domain conditions for composite functions before applying direct substitution.
19. Let $p(x)=\frac{1}{x}+2$. Evaluate $\lim_{x\to 5} p(x)$ using limit laws.
    Answer: $\frac{11}{5}$ This function combines a rational term with a constant. Using the sum rule: $\lim_{x\to 5} p(x) = \lim_{x\to 5} \frac{1}{x} + \lim_{x\to 5} 2 = \frac{1}{5} + 2 = \frac{1}{5} + \frac{10}{5} = \frac{11}{5}$. The rational function $\frac{1}{x}$ is continuous at $x = 5$ since the denominator is non-zero. A common mistake would be incorrectly adding the fractions or not finding a common denominator. The transferable approach is to evaluate each term separately using limit laws, then combine the results using appropriate arithmetic operations.
20. If $h(x)=5-3\cos(x)$ models a signal, evaluate $\lim_{x\to 0} h(x)$.
    Answer: 2 This function involves the cosine function, which is continuous everywhere. Using limit laws and direct substitution: $\lim_{x\to 0} h(x) = 5 - 3\cos(0) = 5 - 3(1) = 5 - 3 = 2$. The key insight is that $\cos(0) = 1$, and we apply the difference rule and scalar multiplication rule for limits. A common mistake would be forgetting that $\cos(0) = 1$ or incorrectly applying the arithmetic operations. The transferable strategy is to memorize key trigonometric values and recognize that trigonometric functions are continuous at standard points, allowing direct substitution.

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