Calculus 1 Quiz: Selecting Limit Procedures

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Selecting Limit ProceduresQuestion 1 of 20

If $k$ is a positive constant, which procedure is necessary to evaluate $\lim_{x \to k} \frac{x-k}{\sqrt{x}-\sqrt{k}}$ ?

Apply L'Hôpital's Rule since direct substitution gives

0/0

Use direct substitution since the function is continuous at

x = k

Analyze the one-sided limits since the square root creates a discontinuity.

Use the Squeeze Theorem since the function involves square roots.

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Calculus 1 Quiz

Calculus 1 Quiz: Selecting Limit Procedures

Practice Selecting Limit Procedures in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Selecting Limit Procedures, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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.

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Question 1

If $k$ is a positive constant, which procedure is necessary to evaluate $\lim_{x \to k} \frac{x-k}{\sqrt{x}-\sqrt{k}}$ ?

Apply L'Hôpital's Rule since direct substitution gives $0/0$ . (correct answer)
Use direct substitution since the function is continuous at $x = k$ .
Analyze the one-sided limits since the square root creates a discontinuity.
Use the Squeeze Theorem since the function involves square roots.

Explanation: Direct substitution yields the indeterminate form

0/0

. For this form, L'Hôpital's Rule is a valid procedure. Applying L'Hôpital's Rule:

\lim_{x \to k} \frac{1}{\frac{1}{2\sqrt{x}}} = \lim_{x \to k} 2\sqrt{x} = 2\sqrt{k}

Question 2

Which statement provides a valid initial procedure and justification for evaluating $\lim_{x \to 3} \frac{x^3 - 27}{x^2 - 9}$ ?

Because the function is a rational expression, divide all terms by $x^3$ .
Because direct substitution fails, factor both polynomials and cancel the common term. (correct answer)
Because the denominator approaches zero, the limit must not exist.
Because the function contains polynomials, direct substitution is sufficient.

Explanation: Direct substitution gives

0/0

, an indeterminate form. For rational functions, this indicates a common factor in the numerator and denominator. Factoring gives

\lim_{x \to 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}

. Canceling the

(x-3)

term is a valid and effective procedure. Applying L'Hôpital's Rule is also valid, but factoring is often considered more fundamental for polynomials.

Question 3

For which of the following limits is L'Hôpital's Rule a valid and effective initial procedure?

$\lim_{x \to \infty} \frac{x + \cos(x)}{x - \cos(x)}$
$\lim_{x \to 2} \frac{x^2 - 4}{x+2}$
$\lim_{x \to 0} \frac{\arctan(x)}{x}$ (correct answer)
$\lim_{x \to 0} \frac{\sin(x)}{x+\cos(x)}$

Explanation: We check the conditions for each option. (A) Applying L'Hôpital's Rule gives

\lim (1-\sin x)/(1+\sin x)

, which does not exist. (B) Direct substitution gives

0/4=0

; the rule is not applicable. (C) Direct substitution gives

0/0

, so the rule is applicable. The derivative is

\lim (1/(1+x^2))/1 = 1

, so it is effective. (D) Direct substitution gives

0/1=0

; the rule is not applicable.

Question 4

To evaluate the limit $L = \lim_{x\to 0^+} (1 + 2x)^{1/x}$ , which of the following describes the necessary initial procedure?

Apply L'Hôpital's Rule directly to the function as it is not a rational expression.
Rewrite the expression as $e^{\ln((1+2x)^{1/x})}$ and then evaluate the limit of the exponent. (correct answer)
Substitute a small positive number for $x$ to approximate the limit's value.
Recognize the indeterminate form $1^\infty$ , which always evaluates to 1.

Explanation: The limit is of the indeterminate form

1^\infty

. This form cannot be evaluated directly, and L'Hôpital's rule cannot be applied to it. The standard procedure is to use the property

A = e^{\ln(A)}

. Let

y = (1 + 2x)^{1/x}

. Then

\ln(y) = \frac{1}{x} \ln(1+2x)

. We evaluate

\lim_{x\to 0^+} \frac{\ln(1+2x)}{x}

, which is of the form

\frac{0}{0}

. Applying L'Hôpital's Rule to the exponent's limit gives

\lim_{x\to 0^+} \frac{2/(1+2x)}{1} = 2

. Since the limit of the exponent is 2, the original limit is

e^2

. Step B describes this correct initial process.

Question 5

Consider the limit $L = \lim_{x\to 0} x^2 \cos(\frac{\pi}{x})$ . Direct substitution yields an indeterminate form. Which of the following procedures is the most appropriate for rigorously evaluating this limit?

Apply L'Hôpital's Rule after rewriting the expression as $\frac{\cos(\pi/x)}{1/x^2}$ .
Use the Squeeze Theorem by bounding the $\cos(\frac{\pi}{x})$ term. (correct answer)
Use the substitution $u = \pi/x$ and apply the known result that $\lim_{u\to\infty} \cos(u)$ does not exist.
Conclude the limit does not exist because $\cos(\frac{\pi}{x})$ oscillates infinitely as $x$ approaches 0.

Explanation: The limit is of the form

0 \cdot \text{(oscillating)}

, which is an indeterminate form. While the cosine term oscillates, the

x^2

term approaches 0 and may dampen the oscillation. L'Hôpital's Rule is not a good choice as the derivative of the rewritten expression is complex and its limit also does not exist. The most appropriate method is the Squeeze Theorem. We know that

-1 \le \cos(\frac{\pi}{x}) \le 1

for all

x \neq 0

. Multiplying the inequality by

x^2

(which is non-negative) gives

-x^2 \le x^2 \cos(\frac{\pi}{x}) \le x^2

. Since

\lim_{x\to 0} (-x^2) = 0

and

\lim_{x\to 0} (x^2) = 0

, by the Squeeze Theorem,

\lim_{x\to 0} x^2 \cos(\frac{\pi}{x}) = 0

Question 6

Which sequence of procedures is most appropriate for evaluating $\lim_{x\to 0} \frac{e^x - 1 - \sin(x)}{1 - \cos(x)}$ ?

Apply L'Hôpital's Rule twice, as the indeterminate form $\frac{0}{0}$ persists after the first application. (correct answer)
Multiply by the conjugate of the denominator, then simplify using known trigonometric limits.
Apply L'Hôpital's Rule once, then use the special limit $\lim_{x\to 0} \frac{\sin(x)}{x} = 1$ .
Separate the expression into two fractions, $\frac{e^x - 1}{1 - \cos(x)} - \frac{\sin(x)}{1 - \cos(x)}$ , and evaluate each limit.

Explanation: When evaluating limits that result in indeterminate forms like

\frac{0}{0}

, you need to systematically work through the algebra or apply techniques like L'Hôpital's Rule until you reach a determinate form. Let's verify this is indeed

\frac{0}{0}

: as

x \to 0

, both

e^x - 1 - \sin(x) \to 0

and

1 - \cos(x) \to 0

. Applying L'Hôpital's Rule once gives us:

\lim_{x\to 0} \frac{e^x - \cos(x)}{\sin(x)}

Checking again: as

x \to 0

, the numerator approaches

1 - 1 = 0

and the denominator approaches

0

, so we still have

\frac{0}{0}

. Applying L'Hôpital's Rule a second time:

\lim_{x\to 0} \frac{e^x + \sin(x)}{\cos(x)} = \frac{1 + 0}{1} = 1

Therefore, A is correct - you need L'Hôpital's Rule twice because the indeterminate form persists after the first application. B is incorrect because multiplying by the conjugate

(1 + \cos(x))

would create

\sin^2(x)

in the denominator, which doesn't simplify the original expression effectively. C fails because after one application of L'Hôpital's Rule, you get

\frac{e^x - \cos(x)}{\sin(x)}

, which is still

\frac{0}{0}

and cannot be resolved using just the standard sine limit. D is problematic because when you separate the fractions, each individual limit

\frac{e^x - 1}{1 - \cos(x)}

and

\frac{\sin(x)}{1 - \cos(x)}

doesn't exist in elementary form, making this approach unnecessarily complex. Study tip: When L'Hôpital's Rule gives you another indeterminate form, don't hesitate to apply it again - some limits require multiple applications to resolve.

Question 7

Let $f$ and $g$ be differentiable functions such that $\lim_{x\to a} f(x) = \infty$ , $\lim_{x\to a} g(x) = \infty$ , and $\lim_{x\to a} \frac{f'(x)}{g'(x)} = L$ . What procedure is justified by these conditions to find $\lim_{x\to a} \frac{f(x)}{g(x)}$ ?

The Squeeze Theorem, by finding two functions that bound the ratio $\frac{f(x)}{g(x)}$ and both approach $L$ .
Direct substitution, which shows the limit is $\frac{\infty}{\infty}$ and is therefore equal to 1.
Factoring out the dominant term from $f(x)$ and $g(x)$ and then canceling common factors.
L'Hôpital's Rule, because the limit is of the indeterminate form $\frac{\infty}{\infty}$ and the limit of the ratio of derivatives exists. (correct answer)

Explanation: The given conditions are the exact premises required to apply L'Hôpital's Rule for the indeterminate form

\frac{\infty}{\infty}

. The rule states that if

\lim_{x\to a} f(x) = \pm \infty

and

\lim_{x\to a} g(x) = \pm \infty

, and if

\lim_{x\to a} \frac{f'(x)}{g'(x)}

exists (and equals

L

), then

\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)} = L

. Therefore, L'Hôpital's Rule is the justified procedure.

Question 8

Which procedure should be used to initiate the evaluation of $\lim_{x\to 1} \left( \frac{1}{x-1} - \frac{2}{x^2-1} \right)$ ?

Combine the two terms into a single fraction by finding a common denominator. (correct answer)
Recognize the form as $\infty - \infty$ and conclude that the limit must be 0.
Apply L'Hôpital's Rule to each fraction separately before subtracting the results.
Use the Squeeze Theorem by bounding each term separately.

Explanation: When you encounter a limit involving the subtraction of two fractions that both approach infinity, you're dealing with an indeterminate form that requires algebraic manipulation before applying limit techniques. Let's see why combining into a single fraction works. First, factor the denominator of the second fraction:

x^2-1 = (x-1)(x+1)

. Now find a common denominator:

\frac{1}{x-1} - \frac{2}{x^2-1} = \frac{1}{x-1} - \frac{2}{(x-1)(x+1)} = \frac{x+1}{(x-1)(x+1)} - \frac{2}{(x-1)(x+1)} = \frac{x+1-2}{(x-1)(x+1)} = \frac{x-1}{(x-1)(x+1)}

After canceling the common factor

(x-1)

, you get

\frac{1}{x+1}

, and

\lim_{x\to 1} \frac{1}{x+1} = \frac{1}{2}

. Answer choice B is incorrect because while this does create an "

\infty - \infty

" indeterminate form, you cannot conclude the limit equals 0 — indeterminate forms require further work to resolve. Choice C is wrong because L'Hôpital's Rule applies to quotients in indeterminate form, not to individual fractions before combining them. Choice D misapplies the Squeeze Theorem, which is used when you can bound a function between two others with the same limit, not relevant here. Study tip: When you see subtraction of rational functions that both blow up at the same point, always try algebraic simplification first. Combine fractions using common denominators before applying advanced techniques like L'Hôpital's Rule.

Question 9

Let the function $f$ be defined by

For what value of the constant $c$ does $\lim_{x\to 3} f(x)$ exist? To answer this question, which of the following procedures must be followed?

Set the derivatives of the two pieces of the function equal to each other at $x=3$ and solve for $c$ .
Set the value of the function from the left, $f(3)$ , equal to the value of the function from the right, $f(3)$ , and solve for $c$ .
Evaluate the left-hand limit and the right-hand limit at $x=3$ and set them equal to each other. (correct answer)
Check for which value of $c$ the function is defined at $x=3$ , as this is sufficient for the limit to exist.

Explanation: For a limit to exist at a point where a function definition changes, the left-hand limit must equal the right-hand limit. The procedure is to calculate both one-sided limits and set them equal. The left-hand limit is

\lim_{x\to 3^-} (cx^2 + 2x) = c(3)^2 + 2(3) = 9c + 6

. The right-hand limit is

\lim_{x\to 3^+} (x^3 - cx) = 3^3 - c(3) = 27 - 3c

. Setting them equal:

9c + 6 = 27 - 3c

, which gives

12c = 21

, so

c = 7/4

. Setting the derivatives equal (Choice A) is the condition for differentiability, not the existence of a limit. Choice B is imprecise; we must use limits, not just function values. Choice D is incorrect; being defined at a point does not guarantee a limit exists.

Question 10

The limit $L = \lim_{h\to 0} \frac{\tan(\pi/4 + h) - 1}{h}$ can be evaluated using several methods. Which procedure provides the most direct evaluation by recognizing the limit's fundamental structure?

Applying L'Hôpital's Rule by differentiating the numerator and denominator with respect to $h$ .
Identifying the limit as the definition of the derivative for a specific function at a specific point. (correct answer)
Using the trigonometric identity for $\tan(A+B)$ to expand the numerator before simplifying.
Multiplying the numerator and denominator by the conjugate, $\tan(\pi/4 + h) + 1$ .

Explanation: The expression is in the form

\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}

, which is the definition of the derivative

f'(a)

. Here,

f(x) = \tan(x)

and

a = \pi/4

. Note that

f(\pi/4) = \tan(\pi/4) = 1

. Therefore, the limit is equivalent to finding the derivative of

\tan(x)

x=\pi/4

. Since

f'(x) = \sec^2(x)

, the limit is

\sec^2(\pi/4) = (\sqrt{2})^2 = 2

. While L'Hôpital's Rule (A) and using trig identities (C) are also valid procedures that yield the correct answer, recognizing the definition of the derivative is the most direct conceptual approach.

Question 11

Consider the limit $L = \lim_{x\to\infty} \frac{x + \sin(x)}{2x + \cos(x)}$ . A student attempts to use L'Hôpital's Rule because the limit appears to be of the indeterminate form $\frac{\infty}{\infty}$ . Which statement best explains why this approach is problematic and identifies a more reliable procedure?

L'Hôpital's Rule is not applicable because the expression is not an indeterminate form as $x \to \infty$ .
The limit does not exist because the numerator and the denominator oscillate indefinitely due to the sine and cosine terms.
Applying L'Hôpital's Rule leads to a limit that does not exist, so the rule is inconclusive; dividing the numerator and denominator by $x$ is the appropriate method. (correct answer)
The Squeeze Theorem must be used, but it fails because the function cannot be bounded by two functions with the same limit.

Explanation: Direct substitution leads to the indeterminate form

\frac{\infty}{\infty}

, so L'Hôpital's Rule seems applicable. Applying the rule gives

\lim_{x\to\infty} \frac{1 + \cos(x)}{2 - \sin(x)}

. This new limit does not exist because the numerator oscillates between 0 and 2, and the denominator oscillates between 1 and 3. When the limit of the ratio of derivatives does not exist, L'Hôpital's Rule is inconclusive. A more reliable procedure is to divide the numerator and denominator by the highest power of

x

\lim_{x\to\infty} \frac{1 + \frac{\sin(x)}{x}}{2 + \frac{\cos(x)}{x}}

. Since

\lim_{x\to\infty} \frac{\sin(x)}{x} = 0

and

\lim_{x\to\infty} \frac{\cos(x)}{x} = 0

, the limit is

\frac{1+0}{2+0} = \frac{1}{2}

Question 12

Consider $\lim_{x \to \infty} \frac{3^x}{x^3}$ . Which procedure is appropriate?

Divide the numerator and denominator by $x^3$ .
Take the natural logarithm of the expression to simplify the exponent.
Conclude the limit is 1 because both functions grow towards infinity.
Apply L'Hôpital's Rule three times. (correct answer)

Explanation: When you encounter limits involving exponential and polynomial functions, you're dealing with a classic indeterminate form that requires careful analysis of growth rates. This limit has the form

\frac{\infty}{\infty}

since both

3^x

and

x^3

approach infinity as

x \to \infty

. This indeterminate form is perfect for L'Hôpital's Rule, which states that if

\lim_{x \to a} \frac{f(x)}{g(x)}

gives

\frac{\infty}{\infty}

, then the limit equals

\lim_{x \to a} \frac{f'(x)}{g'(x)}

(if this limit exists). Applying L'Hôpital's Rule: First application gives

\frac{3^x \ln 3}{3x^2}

, still

\frac{\infty}{\infty}

. Second application:

\frac{3^x (\ln 3)^2}{6x}

, still indeterminate. Third application:

\frac{3^x (\ln 3)^3}{6} = \infty

. Answer D is correct. Option A fails because dividing by

x^3

gives

\frac{(3/x)^x \cdot 3^x}{x^3} \cdot \frac{1/x^3}{1/x^3}

, which doesn't simplify the exponential growth issue. Option B incorrectly suggests taking the logarithm of the entire expression, but we need to evaluate the limit of the original expression, not its logarithm. Option C shows a fundamental misunderstanding—when both numerator and denominator approach infinity, you can't conclude the limit is 1; exponential functions grow much faster than polynomials. Remember: exponential functions always dominate polynomial functions as

x \to \infty

. When you see exponential over polynomial, expect L'Hôpital's Rule and prepare to apply it multiple times.

Question 13

Let $f(x)$ be a differentiable function. A student needs to evaluate $L = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}$ .

Which procedure correctly evaluates this limit in terms of $f(x)$ ?

Use direct substitution by setting $h=0$ , which shows the limit is $0/0$ and therefore equals 1.
Apply L'Hôpital's Rule with respect to $h$ to the expression. (correct answer)
Multiply by the conjugate of the numerator, $f(x+h) + f(x-h)$ .
Conclude the limit cannot be found without an explicit function for $f(x)$ .

Explanation: The limit is of the form

0/0

(assuming continuity from differentiability). The most direct procedure is to apply L'Hôpital's Rule, differentiating the numerator and denominator with respect to the variable

h

. This gives

\lim_{h \to 0} \frac{f'(x+h) \cdot 1 - f'(x-h) \cdot (-1)}{2} = \frac{f'(x) + f'(x)}{2} = f'(x)

. This expression is a known alternative form of the derivative.

Question 14

A function $f(x)$ satisfies the inequality $e^x - 1 \le f(x) \le x^2+x$ for all $x$ near 0.

Which procedure must be used to determine $\lim_{x \to 0} f(x)$ ?

Find the derivative $f'(x)$ and use it to determine the limit.
Evaluate the limits of the bounding functions and apply the Squeeze Theorem. (correct answer)
Assume $f(x)$ is the average of the two bounding functions.
Conclude the limit cannot be determined without an explicit formula for $f(x)$ .

Explanation: The problem provides upper and lower bounds for

f(x)

. The appropriate procedure for this situation is the Squeeze Theorem. We evaluate the limits of the bounds:

\lim_{x \to 0} (e^x - 1) = 1 - 1 = 0

and

\lim_{x \to 0} (x^2+x) = 0

. Since both limits are equal, the Squeeze Theorem guarantees that

\lim_{x \to 0} f(x) = 0

Question 15

To evaluate $\lim_{x \to 0^+} (\cot x)^{\sin x}$ , which sequence of procedures is most appropriate?

Rewrite as $\lim_{x \to 0^+} \frac{\sin x}{\tan x}$ and apply L'Hôpital's Rule.
Use direct substitution, yielding $\infty^0$ , which equals 1.
Let $y = (\cot x)^{\sin x}$ , take the natural logarithm, and transform the resulting product into a quotient. (correct answer)
Apply L'Hôpital's Rule directly to the function as given.

Explanation: The limit has the indeterminate form

\infty^0

. To resolve this, we must use logarithms. Let

y = (\cot x)^{\sin x}

. Then

\ln y = \sin x \ln(\cot x)

. This is a

0 \cdot \infty

form, which must be rewritten as a quotient (e.g.,

\frac{\ln(\cot x)}{\csc x}

) before L'Hôpital's Rule can be applied.

Question 16

To evaluate $\lim_{x\to\infty} \frac{3e^{2x} + 5e^{-x}}{4e^{2x} - 2e^{-x}}$ , what is the most effective initial step?

Apply L'Hôpital's Rule, as the limit is of the form $\frac{\infty}{\infty}$ .
Divide both the numerator and the denominator by the fastest-growing term, $e^{2x}$ . (correct answer)
Find a common denominator for the terms in the expression.
Substitute $u=e^x$ and analyze the resulting rational function as $u \to \infty$ .

Explanation: As

x \to \infty

, the term

e^{2x}

grows much faster than

e^{-x}

(which approaches 0). This makes

e^{2x}

the dominant term. The most effective procedure for limits at infinity involving such expressions is to divide the numerator and the denominator by the highest-growth term. This yields

\lim_{x\to\infty} \frac{3 + 5e^{-3x}}{4 - 2e^{-3x}}

. As

x \to \infty

e^{-3x} \to 0

, so the limit is

\frac{3+0}{4-0} = \frac{3}{4}

. While L'Hôpital's rule is applicable, it would result in a similar expression and would need to be applied indefinitely, making it ineffective. The substitution in choice D is valid but ultimately leads back to the same analysis as choice B.

Question 17

To correctly evaluate $\lim_{x\to 2} \frac{x^2 - 4}{|x - 2|}$ , which of the following procedures is essential?

Evaluate the left-hand and right-hand limits separately due to the presence of the absolute value function. (correct answer)
Simplify the expression by canceling a factor of $(x-2)$ from the numerator and denominator.
Apply L'Hôpital's Rule, as the limit is of the form $\frac{0}{0}$ .
Multiply the numerator and denominator by the conjugate of the denominator, which is $|x-2|$ .

Explanation: When you encounter a limit involving an absolute value function, you must always consider what happens on both sides of the point where the expression inside the absolute value equals zero. Here,

|x-2|

changes its behavior at

x = 2

, so you cannot treat this as a standard limit problem. The correct approach (A) requires evaluating left-hand and right-hand limits separately. For

x < 2

, we have

|x-2| = -(x-2) = 2-x

, so the limit becomes

\lim_{x \to 2^-} \frac{x^2-4}{2-x} = \lim_{x \to 2^-} \frac{(x-2)(x+2)}{-(x-2)} = -4

. For

x > 2

, we have

|x-2| = x-2

, so

\lim_{x \to 2^+} \frac{x^2-4}{x-2} = \lim_{x \to 2^+} \frac{(x-2)(x+2)}{x-2} = 4

. Since the one-sided limits differ, the limit does not exist. Option B fails because you cannot directly cancel

(x-2)

when the denominator contains

|x-2|

—the absolute value changes the sign depending on which side of 2 you approach from. Option C is incorrect because L'Hôpital's Rule doesn't resolve the fundamental issue that the absolute value creates different behaviors on each side. Option D makes no sense mathematically—

|x-2|

is already simplified and multiplying by it would only complicate the expression. Remember: whenever you see an absolute value in a limit problem, immediately check if the expression inside equals zero at your approach point. If so, evaluate one-sided limits separately.

Question 18

A student needs to find the limit $L = \lim_{x\to\infty} (\sqrt{x^2 + 6x} - x)$ . Which of the following procedures must be applied to transform the expression from its initial indeterminate form?

Apply L'Hôpital's Rule to the expression as it is written.
Divide all terms by the highest power of $x$ present in the expression.
Multiply the expression by its conjugate, $\frac{\sqrt{x^2 + 6x} + x}{\sqrt{x^2 + 6x} + x}$ . (correct answer)
Factor out an $x$ from the expression to get $x(\sqrt{1 + 6/x} - 1)$ and evaluate the limit of each factor.

Explanation: The limit is of the indeterminate form

\infty - \infty

. L'Hôpital's Rule cannot be applied directly because the expression is not a fraction. Dividing by

x

(Choice B) leads to

\sqrt{1+6/x}-1

, which approaches 0, an incorrect result. Factoring out

x

(Choice D) leads to the indeterminate form

\infty \cdot 0

, which still requires manipulation. The correct initial procedure is to multiply by the conjugate to transform the expression into a rational form. This gives

\lim_{x\to\infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x\to\infty} \frac{6x}{\sqrt{x^2 + 6x} + x}

. This is now in the

\frac{\infty}{\infty}

form, and the limit can be found by dividing the numerator and denominator by

x

, yielding

\frac{6}{\sqrt{1+6/x}+1} = \frac{6}{1+1} = 3

Question 19

To evaluate the limit $\lim_{x \to \infty} (\sqrt{4x^2 + 5x} - 2x)$ , which of the following is the most appropriate initial step?

Apply L'Hôpital's Rule directly to the expression.
Multiply the expression by its conjugate, $\frac{\sqrt{4x^2 + 5x} + 2x}{\sqrt{4x^2 + 5x} + 2x}$ . (correct answer)
Divide each term by the highest power of $x$ , which is $x^2$ .
Use direct substitution, which shows the limit approaches $\infty - \infty = 0$ .

Explanation: The limit presents the indeterminate form

\infty - \infty

. The standard and most effective procedure for limits of this type involving square roots is to multiply by the conjugate. This will transform the expression into a rational function, which can then be evaluated by dividing by the highest power of

x

Question 20

To evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{2x}$ , which of the following describes the necessary first step?

Use direct substitution to get $(1+0)^\infty$ , which evaluates to 1.
Apply L'Hôpital's Rule directly to the expression as written.
Distribute the exponent $2x$ to each term inside the parentheses.
Set $y$ equal to the expression and evaluate the limit of $\ln(y)$ to handle the indeterminate form. (correct answer)

Explanation: The limit is of the indeterminate form

1^\infty

. This form must be manipulated before a limit can be found. The standard procedure is to let

y = (1 + 3/x)^{2x}

, take the natural logarithm of both sides (

\ln y = 2x \ln(1+3/x)

), and then evaluate

\lim_{x \to \infty} \ln y

. This transforms the problem into a

\infty \cdot 0

form, which can be rearranged for L'Hôpital's Rule.