AP Calculus AB Quiz: Selecting Procedures For Determining Limits

What this quiz covers

This quiz focuses on Selecting Procedures For Determining Limits, 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.

Question 1

1. Multiply the numerator and denominator by $x$ and rearrange to use the known limits for $\frac{e^x-1}{x}$ and $\frac{x}{\sin(x)}$. (correct answer)
2. Multiply the numerator and denominator by the conjugate of the numerator, $e^x + 1$, to simplify the expression.
3. Apply the Squeeze Theorem by bounding the function between two simpler functions that approach the same value.
4. Factor $e^x$ from the numerator and use trigonometric identities to simplify the denominator.

Explanation: 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 $\frac{\lim_{x \to 0} \frac{e^x-1}{x}}{\lim_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1$.

Question 2

1. Multiply the numerator and denominator by the conjugate of the numerator, $\sqrt{9x^2+x}$.
2. Divide the numerator and denominator by $x$, remembering that for $x>0$, $x = \sqrt{x^2}$. (correct answer)
3. Apply the Squeeze Theorem, since the function contains a square root and is difficult to evaluate directly.
4. Use direct substitution, which is the first step for all limit problems and will give the answer here.

Explanation: 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.

Question 3

1. Multiply the numerator and denominator by the conjugate of the numerator.
2. Cancel the term $h$ from the numerator and the denominator immediately.
3. Combine the fractions in the numerator into a single fraction using a common denominator. (correct answer)
4. Apply direct substitution to the expression as it is written to find the final value.

Explanation: 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.

Question 4

1. Because direct substitution results in $\frac{0}{0}$, one should simplify the expression by factoring.
2. The limit must be evaluated by considering the limits from the left and right, as the definition of $|x-1|$ changes at $x=1$. (correct answer)
3. The expression can be simplified by multiplying the numerator and denominator by the conjugate of the denominator.
4. Since the expression contains an absolute value, the Squeeze Theorem must be used to find the limit.

Explanation: 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.

Question 5

1. Evaluating $f(2)$ directly since the function is defined at $x=2$ and using that value as the limit.
2. Evaluating the limit from the left ($x \to 2^-$) and the limit from the right ($x \to 2^+$) and comparing them. (correct answer)
3. Factoring the quadratic and linear expressions and canceling common terms before taking the limit.
4. Applying the Squeeze Theorem because the function is defined by two different rules on each side of $x=2$.

Explanation: 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.

Question 6

1. Factoring the numerator and denominator to cancel common polynomial factors.
2. Multiplying the numerator and denominator by the conjugate of the denominator.
3. Dividing every term in the numerator and denominator by $x^3$, the highest power of $x$ in the denominator. (correct answer)
4. Applying the Squeeze Theorem by comparing the function to simpler bounding functions.

Explanation: 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.

Question 7

1. Multiplying the numerator and denominator by the conjugate of the numerator, $x^2+9$.
2. Factoring the expressions in the numerator and denominator and simplifying the resulting rational expression. (correct answer)
3. Using the Squeeze Theorem with functions that bound the given expression near $x=3$.
4. Applying the special trigonometric limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ after a substitution.

Explanation: 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.

Question 8

1. Use direct substitution into the expression $\frac{1}{x-3}$ because it defines the function near $x=3$.
2. Evaluate the one-sided limits as $x \to 3^-$ and $x \to 3^+$ to determine if the function approaches $\infty$ or $-\infty$. (correct answer)
3. Conclude the limit is 5 because the function is explicitly defined as $g(3)=5$.
4. Factor the denominator and simplify the expression before applying direct substitution.

Explanation: 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.

Question 9

1. Factoring a common term of $x$ from the numerator and denominator.
2. Multiplying the numerator and denominator by the conjugate expression $\sqrt{x+9} + 3$. (correct answer)
3. Applying the Squeeze Theorem by bounding the function between $y=-1/x$ and $y=1/x$.
4. Simplifying the expression by finding a common denominator for terms in the numerator.

Explanation: 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.

Question 10

1. Direct substitution, because the function $f(x) = \frac{\sin(x)}{x}$ is continuous at $x=\pi/2$. (correct answer)
2. Algebraic simplification by factoring, because the expression is a rational function of trigonometric terms.
3. The Squeeze Theorem, because the function involves a trigonometric component which is bounded.
4. Multiplying by a conjugate, which is a standard method for resolving indeterminate forms.

Explanation: 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.

Question 11

1. Direct substitution, as the function consists of products of continuous functions.
2. Algebraic simplification by factoring out $x^2$ and canceling terms.
3. The Squeeze Theorem, by establishing upper and lower bounds for the function. (correct answer)
4. Rationalizing the argument of the cosine function by multiplying by its conjugate.

Explanation: 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.

Question 12

1. Multiply the numerator and denominator by the conjugate, $1 + \cos^2(x)$.
2. Apply the Pythagorean identity $1 - \cos^2(x) = \sin^2(x)$ to the numerator. (correct answer)
3. Use direct substitution, since cosine is a continuous function everywhere.
4. Use the Squeeze Theorem, as the expression involves a trigonometric function.

Explanation: 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 $1^2 = 1$.

Question 13

1. Factor the numerator and denominator to simplify the expression and resolve the indeterminate form.
2. Multiply by the conjugate of the denominator, $x+2$, to remove the zero in the denominator.
3. Conclude that the limit does not exist and investigate the one-sided limits to determine if it is $\infty$ or $-\infty$. (correct answer)
4. Apply the Squeeze Theorem, since algebraic manipulation is not possible to simplify the expression.

Explanation: 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.

Question 14

1. Attempt to factor the expression for $f(x)$ and simplify.
2. Check if the conditions for the Squeeze Theorem can be met.
3. Evaluate the one-sided limits as $x \to a^-$ and $x \to a^+$ separately.
4. Attempt to use direct substitution by evaluating $f(a)$. (correct answer)

Explanation: 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.

Question 15

1. Multiply the entire expression by the conjugate of the first term's denominator.
2. Combine the two terms into a single fraction using a common denominator. (correct answer)
3. Apply the Squeeze Theorem by finding functions that bound each term separately.
4. Factor the term $1/x$ from the expression and evaluate the limits of the factors.

Explanation: 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.

Question 16

1. The function $f(x)$ must be defined and continuous at $x=c$.
2. The functions $g(x)$ and $h(x)$ must be differentiable at $x=c$.
3. The limits of the bounding functions must exist and be equal: $\lim_{x \to c} g(x) = \lim_{x \to c} h(x)$. (correct answer)
4. The function $f(x)$ must involve a trigonometric component like sine or cosine.

Explanation: 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.

Question 17

1. Multiplying the numerator and denominator by the conjugate of the numerator, $x^3 + 8$.
2. Factoring the numerator as a difference of cubes using the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$. (correct answer)
3. Dividing the numerator and denominator by the highest power of $x$, which is $x^3$.
4. Recognizing the expression as the definition of the derivative of $f(x)=x^3$ at $x=2$.

Explanation: 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.

Question 18

1. Apply direct substitution into the expression as it is written, as both functions are continuous.
2. Use the Squeeze Theorem, since both numerator and denominator involve bounded trigonometric functions.
3. Rewrite $\cot(x)$ as $\frac{\cos(x)}{\sin(x)}$ and simplify the resulting complex fraction. (correct answer)
4. Multiply the numerator and denominator by $\sin(x)$ to eliminate the cotangent function.

Explanation: 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.

Question 19

1. Recognize the expression as the definition of the derivative of $f(x)=\ln(x)$ at $x=1$. (correct answer)
2. Since the form is $\frac{0}{0}$, the expression must be simplified by factoring the numerator and denominator.
3. Since the form is $\frac{0}{0}$, the numerator and denominator should be multiplied by their conjugates.
4. This limit represents a removable discontinuity that can be resolved by algebraic simplification.

Explanation: 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.