Selecting Procedures for Determining Limits - AP Calculus AB

1. To analyze the limit of the function $g(x)$ as $x$ approaches 3, which procedure is necessary?
   Answer: Evaluate the one-sided limits as $x \to 3^-$ and $x \to 3^+$ to determine if the function approaches $\infty$ or $-\infty$. The limit of a function at a point depends on the values of the function near that point, not at the point. Near $x=3$, the function is defined by $\frac{1}{x-3}$. Substituting $x=3$ into this expression gives the form $\frac{1}{0}$, which indicates a vertical asymptote. To fully describe the behavior, it is necessary to examine the one-sided limits. The value $g(3)=5$ is irrelevant to the value of the limit.
2. Which of the following theorems or methods is most appropriate for evaluating the limit $\lim_{x \to 0} x^2 \cos\left(\frac{1}{x^2}\right)$?
   Answer: The Squeeze Theorem, by establishing upper and lower bounds for the function. The term $\cos(1/x^2)$ oscillates infinitely often between -1 and 1 as $x$ approaches 0, so the limit cannot be found by direct substitution. The Squeeze Theorem is the appropriate method. Since $-1 \le \cos(1/x^2) \le 1$, we can write $-x^2 \le x^2 \cos(1/x^2) \le x^2$. As $x \to 0$, both $-x^2$ and $x^2$ approach 0, so the given limit must also be 0.
3. A student is asked to find $\lim_{x \to a} f(x)$. What is the first procedure the student should always attempt?
   Answer: Attempt to use direct substitution by evaluating $f(a)$. The first step in evaluating any limit should always be to try direct substitution. If the function is continuous at $x=a$, this will yield the limit's value immediately. If direct substitution results in an indeterminate form (like $\frac{0}{0}$) or indicates a discontinuity (like $\frac{k}{0}$), this initial step informs which of the other procedures is necessary.
4. What is the most effective initial step to simplify the limit expression $\lim_{x \to 0} \frac{1 - \cos^2(x)}{x^2}$ before evaluating it?
   Answer: Apply the Pythagorean identity $1 - \cos^2(x) = \sin^2(x)$ to the numerator. Direct substitution yields the indeterminate form $\frac{0}{0}$. The most direct way to simplify the expression is to use the fundamental Pythagorean identity $1 - \cos^2(x) = \sin^2(x)$. The limit then becomes $\lim_{x \to 0} \frac{\sin^2(x)}{x^2} = \lim_{x \to 0} \left(\frac{\sin(x)}{x}\right)^2$, which can be evaluated using the special trigonometric limit, resulting in $1^2 = 1$.
5. What additional condition is necessary to apply the Squeeze Theorem and reach a conclusion?
   Answer: The limits of the bounding functions must exist and be equal: $\lim_{x \to c} g(x) = \lim_{x \to c} h(x)$. The Squeeze Theorem requires three conditions: (1) $g(x) \le f(x) \le h(x)$ in an interval around $c$, (2) $\lim_{x \to c} g(x) = L$, and (3) $\lim_{x \to c} h(x) = L$. The crucial condition to draw a conclusion about $f(x)$ is that the limits of the two bounding functions must be equal to the same finite value, $L$. If this is true, then $\lim_{x \to c} f(x) = L$ as well.
6. The limit $\lim_{x \to 0} \left( \frac{1}{x(x+1)} - \frac{1}{x} \right)$ results in the indeterminate form $\infty - \infty$. What is the most appropriate first algebraic step to evaluate this limit?
   Answer: Combine the two terms into a single fraction using a common denominator. To resolve the indeterminate form $\infty - \infty$, the standard procedure is to combine the terms into a single rational expression. By finding the common denominator $x(x+1)$, the expression simplifies to $\lim_{x \to 0} \frac{1 - (x+1)}{x(x+1)} = \lim_{x \to 0} \frac{-x}{x(x+1)}$. This creates a new form that can be simplified by cancellation.
7. To evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x+9} - 3}{x}$, evaluating using direct substitution results in an indeterminate form. Which of the following is the most appropriate next step?
   Answer: Multiplying the numerator and denominator by the conjugate expression $\sqrt{x+9} + 3$. The limit results in the indeterminate form $\frac{0}{0}$. The presence of a square root in the numerator suggests that multiplying the numerator and denominator by its conjugate, $\sqrt{x+9} + 3$, is the most effective procedure. This step removes the radical from the numerator and creates a term that can be cancelled with the denominator.
8. Which of the following is the most effective method for determining the limit $\lim_{x \to \infty} \frac{3x^3 - 2x + 1}{5x^3 + x^2 - 7}$?
   Answer: Dividing every term in the numerator and denominator by $x^3$, the highest power of $x$ in the denominator. For limits at infinity of rational functions, the standard and most effective procedure is to analyze the end behavior by dividing both the numerator and the denominator by the highest power of $x$ that appears in the denominator. This transforms the expression into a form where the limits of individual terms can be easily evaluated as $x$ approaches infinity.
9. Which of the following represents the best first step to simplify and evaluate the limit $\lim_{x \to \pi/2} \frac{\cos(x)}{\cot(x)}$?
   Answer: Rewrite $\cot(x)$ as $\frac{\cos(x)}{\sin(x)}$ and simplify the resulting complex fraction. Direct substitution yields the indeterminate form $\frac{0}{0}$. The most direct simplification is to use the identity $\cot(x) = \frac{\cos(x)}{\sin(x)}$. The expression becomes $\frac{\cos(x)}{(\cos(x)/\sin(x))}$. For values of $x$ near $\pi/2$, $\cos(x) \neq 0$, so this simplifies to $\sin(x)$. The limit can then be evaluated by direct substitution.
10. To evaluate the limit $\lim_{x \to 3} \frac{x^2 - 9}{x^2 - 2x - 3}$, direct substitution results in the indeterminate form $\frac{0}{0}$. Which of the following is the most appropriate algebraic method to use next?
    Answer: Factoring the expressions in the numerator and denominator and simplifying the resulting rational expression. The expression is a rational function. For the indeterminate form $\frac{0}{0}$, the appropriate procedure is to factor the numerator into $(x-3)(x+3)$ and the denominator into $(x-3)(x+1)$. This allows for cancellation of the $(x-3)$ term, which resolves the indeterminate form and allows for evaluation by direct substitution into the simplified expression.
11. To evaluate the limit $\lim_{h \to 0} \frac{\frac{1}{4+h} - \frac{1}{4}}{h}$, which of the following algebraic procedures should be performed first?
    Answer: Combine the fractions in the numerator into a single fraction using a common denominator. Direct substitution results in the indeterminate form $\frac{0}{0}$. The expression contains a complex fraction in the numerator. The correct first step is to simplify this by finding a common denominator for the terms $\frac{1}{4+h}$ and $\frac{1}{4}$. This will result in a single fraction in the numerator, which can then be simplified further, allowing for cancellation of the $h$ term in the denominator.
12. Which of the following is the most appropriate method to find the limit $\lim_{x \to \infty} \frac{\sqrt{9x^2 + x}}{2x+1}$?
    Answer: Divide the numerator and denominator by $x$, remembering that for $x>0$, $x = \sqrt{x^2}$. For a limit at infinity involving a rational-like expression with a radical, the most effective method is to divide the numerator and denominator by the highest power of $x$ in the denominator, which is $x$. To divide the numerator by $x$, we use the fact that $x = \sqrt{x^2}$ for positive $x$ to bring the term inside the square root. This simplifies the expression and allows for evaluation of the limit.
13. Which of the following procedures should be used to evaluate the limit $\lim_{x \to \pi/2} \frac{\sin(x)}{x}$?
    Answer: Direct substitution, because the function $f(x) = \frac{\sin(x)}{x}$ is continuous at $x=\pi/2$. The function $f(x) = \frac{\sin(x)}{x}$ is continuous for all $x \neq 0$. Since the limit is being evaluated at $x=\pi/2$, a point within the domain of continuity, direct substitution is the correct and simplest method. The result is $\frac{\sin(\pi/2)}{\pi/2} = \frac{1}{\pi/2} = \frac{2}{\pi}$. No other procedure is necessary.
14. Which of the following statements describes the most appropriate procedure for determining the limit $\lim_{x \to 1} \frac{x-1}{|x-1|}$?
    Answer: The limit must be evaluated by considering the limits from the left and right, as the definition of $|x-1|$ changes at $x=1$. The absolute value function $|x-1|$ is defined piecewise: $|x-1| = x-1$ for $x > 1$ and $|x-1| = -(x-1)$ for $x < 1$. Because the rule changes at $x=1$, it is essential to evaluate the one-sided limits. The limit from the right is 1, and the limit from the left is -1. Since they differ, the two-sided limit does not exist.
15. The limit $\lim_{x \to 2} \frac{x^3 - 8}{x - 2}$ results in the indeterminate form $\frac{0}{0}$. Which algebraic technique is most suitable for evaluating this limit?
    Answer: Factoring the numerator as a difference of cubes using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$. While choice D is also a valid method, the question asks for the most suitable algebraic technique. The expression in the numerator, $x^3 - 8$, is a difference of cubes, which factors into $(x-2)(x^2+2x+4)$. After factoring, the $(x-2)$ term can be canceled from the numerator and denominator, which resolves the indeterminate form and allows for evaluation.
16. To determine $\lim_{x \to 2} f(x)$, which of the following procedures is required?
    Answer: Evaluating the limit from the left ($x \to 2^-$) and the limit from the right ($x \to 2^+$) and comparing them. Because the function's definition changes at the point $x=2$, it is necessary to determine if the limit from the left is equal to the limit from the right. One must calculate $\lim_{x \to 2^-} f(x)$ using $x^2+1$ and $\lim_{x \to 2^+} f(x)$ using $3x-1$. The two-sided limit exists if and only if these one-sided limits are equal.
17. Direct substitution into the limit $\lim_{x \to 0} \frac{e^x - 1}{\sin(x)}$ yields $\frac{0}{0}$. Which procedure is most effective for evaluating this limit without using L'Hôpital's Rule?
    Answer: Multiply the numerator and denominator by $x$ and rearrange to use the known limits for $\frac{e^x-1}{x}$ and $\frac{x}{\sin(x)}$. This limit can be resolved by using two known fundamental limits: $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$. The expression can be rewritten as $\lim_{x \to 0} \frac{(e^x - 1)/x}{\sin(x)/x}$. Using the quotient property of limits, the limit is $\frac{\lim_{x \to 0} \frac{e^x-1}{x}}{\lim_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1$.
18. Consider the limit $\lim_{x \to 1} \frac{\ln(x)}{x-1}$. A student uses direct substitution and finds the form $\frac{0}{0}$. Which of the following describes a valid conceptual procedure to find this limit?
    Answer: Recognize the expression as the definition of the derivative of $f(x)=\ln(x)$ at $x=1$. The expression $\lim_{x \to 1} \frac{\ln(x) - \ln(1)}{x-1}$ (noting that $\ln(1)=0$) is the formal definition of the derivative of the function $f(x)=\ln(x)$ at the point $a=1$. The derivative of $f(x)=\ln(x)$ is $f'(x)=1/x$. Evaluating this derivative at $x=1$ gives $f'(1)=1$. This conceptual shortcut is a valid procedure.
19. When attempting to evaluate the limit $\lim_{x \to 2} \frac{x+5}{x-2}$, direct substitution results in the form $\frac{7}{0}$. What procedure should be followed next?
    Answer: Conclude that the limit does not exist and investigate the one-sided limits to determine if it is $\infty$ or $-\infty$. The form $\frac{k}{0}$ (where $k \neq 0$) is not an indeterminate form. It indicates the presence of a vertical asymptote at $x=2$. Therefore, the finite limit does not exist. The next step is to analyze the sign of the expression as $x$ approaches 2 from the left and from the right to determine if the function tends towards positive or negative infinity.

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