Calculus 1 Quiz: Limits Via Algebraic Properties

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Limits Via Algebraic PropertiesQuestion 1 of 20

Evaluate the limit: $\lim_{x \to 1} \frac{\sqrt{x+3} - 2}{x^2 - x}$

1

1/2

1/4

0

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Calculus 1 Quiz

Calculus 1 Quiz: Limits Via Algebraic Properties

Practice Limits Via Algebraic Properties 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 Limits Via Algebraic Properties, 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.

All questions

Question 1

Evaluate the limit: $\lim_{x \to 1} \frac{\sqrt{x+3} - 2}{x^2 - x}$

$1$
$1/2$
$1/4$ (correct answer)
$0$

Explanation: When you encounter a limit that gives the indeterminate form

\frac{0}{0}

, you need to manipulate the expression algebraically before evaluating. Let's check: as

x \to 1

, the numerator becomes

\sqrt{1+3} - 2 = 2 - 2 = 0

, and the denominator becomes

1^2 - 1 = 0

. This confirms we have the indeterminate form

\frac{0}{0}

. To resolve this, rationalize the numerator by multiplying both numerator and denominator by the conjugate

\sqrt{x+3} + 2

\frac{\sqrt{x+3} - 2}{x^2 - x} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2} = \frac{(x+3) - 4}{(x^2 - x)(\sqrt{x+3} + 2)} = \frac{x - 1}{(x^2 - x)(\sqrt{x+3} + 2)}

Since

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

, we can cancel the common factor

(x-1)

\frac{x - 1}{x(x-1)(\sqrt{x+3} + 2)} = \frac{1}{x(\sqrt{x+3} + 2)}

Now we can safely substitute

x = 1

\lim_{x \to 1} \frac{1}{x(\sqrt{x+3} + 2)} = \frac{1}{1(\sqrt{4} + 2)} = \frac{1}{1(2 + 2)} = \frac{1}{4}

Answer choice A) 1 would result from incorrectly canceling without proper rationalization. Answer choice B)

\frac{1}{2}

likely comes from an error in the conjugate multiplication or forgetting the factor of

x

in the denominator. Answer choice D) 0 would be the mistake of thinking the limit doesn't exist because of the initial

\frac{0}{0}

form. Strategy tip: When you see square roots in limits that produce

\frac{0}{0}

, rationalization using the conjugate is almost always your first move. Always factor denominators completely to identify potential cancellations.

Question 2

Evaluate the limit: $\lim_{x \to 0} \frac{(x+2)^{-1} - 2^{-1}}{x}$

Does not exist
$-\frac{1}{2}$
$\frac{1}{4}$
$-\frac{1}{4}$ (correct answer)

Explanation: The limit is of the indeterminate form

\frac{0}{0}

. First, rewrite the negative exponents as fractions:

\lim_{x \to 0} \frac{\frac{1}{x+2} - \frac{1}{2}}{x}

Combine the fractions in the numerator by finding a common denominator, which is

2(x+2)

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

Now, simplify the complex fraction:

\lim_{x \to 0} \frac{-x}{2x(x+2)}

Cancel the

x

terms:

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

Substitute

x=0

\frac{-1}{2(0+2)} = -\frac{1}{4}

A: A common guess for limits that initially appear to have a zero in the denominator. B: This can result from forgetting the '2' from the

\frac{1}{2}

term in the common denominator. C: This choice indicates a sign error.

Question 3

Given that $\lim_{x \to 2} \frac{f(x)-5}{x-2} = 3$ , what is $\lim_{x \to 2} f(x)$ ?

0
2
3
5 (correct answer)

Explanation: The given equation states that the limit of a quotient exists and is finite. Since the denominator

(x-2)

approaches 0 as

x \to 2

, the numerator must also approach 0 for the limit to be a finite number. If the numerator approached a non-zero constant, the limit would be infinite. Therefore, we must have:

\lim_{x \to 2} (f(x) - 5) = 0

Using the difference rule for limits:

\lim_{x \to 2} f(x) - \lim_{x \to 2} 5 = 0

\lim_{x \to 2} f(x) - 5 = 0

Solving for the desired limit:

\lim_{x \to 2} f(x) = 5

A: This is the required limit of the numerator,

\lim(f(x)-5)

, not the limit of

f(x)

itself. B: This is the value that

x

is approaching. C: This is the value of the given limit (which represents the derivative

f'(2)

), not the value of

\lim f(x)

Question 4

Let $f(x) = x-2$ and $g(x) = \frac{1}{x-2}$ . What is $\lim_{x \to 2} (f(x) \cdot g(x))$ ?

$\infty$
$0$
Does Not Exist
$1$ (correct answer)

Explanation: When you encounter limits involving products where one factor approaches zero and another approaches infinity, you need to simplify the expression first rather than trying to evaluate each factor separately. Let's compute the product

f(x) \cdot g(x) = (x-2) \cdot \frac{1}{x-2}

. For any value of

x

where

x \neq 2

, we can simplify this expression by canceling the

(x-2)

terms:

(x-2) \cdot \frac{1}{x-2} = 1

Since the simplified expression equals 1 for all

x \neq 2

, we have

\lim_{x \to 2} (f(x) \cdot g(x)) = \lim_{x \to 2} 1 = 1

This makes D correct. A is wrong because even though

g(x) = \frac{1}{x-2}

approaches infinity as

x

approaches 2, the product doesn't blow up due to the cancellation with the

(x-2)

factor from

f(x)

. B is wrong because even though

f(x) = x-2

approaches zero as

x

approaches 2, this creates a

0 \cdot \infty

indeterminate form, not a limit of 0. You must resolve the indeterminate form first. C is wrong because the limit does exist after simplification. The fact that we have an indeterminate form initially doesn't mean the limit doesn't exist—it means we need to work harder to find it. Study tip: When you see indeterminate forms like

0 \cdot \infty

, always try algebraic simplification first. Look for common factors that cancel before attempting more advanced techniques like L'Hôpital's rule.

Question 5

Given $\lim_{x \to 5} f(x) = 10$ and $\lim_{x \to 5} g(x) = 0$ . Additionally, it is known that $g(x) > 0$ for all $x$ in the interval $(4, 6)$ . Which of the following must be true about $\lim_{x \to 5} \frac{f(x)}{g(x)}$ ?

The limit is $-\infty$ .
The limit is $0$ .
The limit is $+\infty$ . (correct answer)
The limit cannot be determined from the given information.

Explanation: When you encounter limits involving division where the denominator approaches zero, you're dealing with a potential infinite limit situation. The key is analyzing the signs and behaviors of both the numerator and denominator. Here,

\lim_{x \to 5} f(x) = 10

means

f(x)

approaches the positive value 10. Meanwhile,

\lim_{x \to 5} g(x) = 0

with

g(x) > 0

in the interval

(4,6)

means

g(x)

approaches 0 from the positive side (since

g(x)

is always positive near

x = 5

). When you divide a number approaching a positive constant by a number approaching zero from the positive side, the quotient grows without bound toward positive infinity. Think of it as

\frac{10}{\text{very small positive number}}

- this gives you a very large positive result. As

g(x)

gets closer to zero while staying positive,

\frac{f(x)}{g(x)}

becomes arbitrarily large and positive. Choice A is wrong because we get positive infinity, not negative infinity, since both

f(x)

and

g(x)

approach positive values. Choice B incorrectly suggests the limit is finite - but dividing by values approaching zero typically creates infinite limits, not zero. Choice D is incorrect because we have sufficient information: knowing the signs and limiting behaviors of both functions completely determines the behavior of their quotient. Remember: when the numerator approaches a non-zero constant and the denominator approaches zero, always check the sign of the denominator near the limit point. The sign determines whether you get

+\infty

-\infty

Question 6

Given $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ , where $L$ and $M$ are finite real numbers. Under which condition can we NOT conclude that $\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}$ ?

$L = 0$ and $M \neq 0$
$M = 0$ (correct answer)
$L \neq 0$ and $M \neq 0$
$f(c)$ or $g(c)$ is undefined

Explanation: The quotient rule for limits states that if

\lim_{x \to c} f(x) = L

and

\lim_{x \to c} g(x) = M

both exist as finite numbers, then

\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{L}{M}

, provided that

M \neq 0

. If

M=0

, the quotient rule cannot be applied, and the limit may be infinite or may require further analysis if it is of the indeterminate form

0/0

. Therefore, the conclusion is not guaranteed when

M=0

. The values of

f(c)

and

g(c)

are irrelevant to the value of the limit as

x

approaches

c

Question 7

Evaluate the limit: $\lim_{x \to 3^-} \frac{x^2 - 9}{|x-3|}$

Does Not Exist
$6$
$0$
$-6$ (correct answer)

Explanation: When evaluating one-sided limits involving absolute values, you need to carefully consider what happens to the absolute value expression as you approach from the specified direction. Since we're approaching 3 from the left (

x \to 3^-

), we have

x < 3

, which means

x - 3 < 0

. This is crucial because when the expression inside absolute value bars is negative,

|x-3| = -(x-3) = 3-x

. Now we can rewrite the limit:

\lim_{x \to 3^-} \frac{x^2 - 9}{|x-3|} = \lim_{x \to 3^-} \frac{x^2 - 9}{3-x}

Factor the numerator:

x^2 - 9 = (x-3)(x+3)

So our limit becomes:

\lim_{x \to 3^-} \frac{(x-3)(x+3)}{3-x} = \lim_{x \to 3^-} \frac{(x-3)(x+3)}{-(x-3)} = \lim_{x \to 3^-} \frac{(x+3)}{-1} = \lim_{x \to 3^-} -(x+3)

x

approaches 3, this equals

-(3+3) = -6

. Answer choice (D)

-6

is correct. Choice (B)

6

would be the result if you approached from the right or forgot the negative sign from the absolute value. Choice (C)

0

might tempt you if you incorrectly thought the

(x-3)

factors simply canceled to give zero. Choice (A) "Does Not Exist" is wrong because the one-sided limit does exist and equals

-6

. Key strategy: Always determine the sign of the expression inside absolute value bars based on your approach direction before simplifying.

Question 8

Let $f(x) = \frac{2x^2 + ax + b}{x-3}$ . If $\lim_{x \to 3} f(x) = 10$ , what is the value of the sum $a+b$ ?

-2
-12
-14 (correct answer)
-22

Explanation: For the limit of the rational function to exist as

x \to 3

, while the denominator approaches 0, the numerator must also approach 0. This creates an indeterminate form that can be simplified. First, set the numerator to 0 at

x=3

2(3)^2 + a(3) + b = 0 \implies 18 + 3a + b = 0

This means

(x-3)

must be a factor of the numerator. We can use this fact to find the limit in terms of

a

\lim_{x \to 3} \frac{2x^2 + ax + b}{x-3} = 10

Using L'Hôpital's Rule (or by factoring), we differentiate the numerator and denominator:

\lim_{x \to 3} \frac{4x+a}{1} = 10

Substitute

x=3

4(3) + a = 10 \implies 12 + a = 10 \implies a = -2

Now substitute

a=-2

back into the first equation:

18 + 3(-2) + b = 0 \implies 18 - 6 + b = 0 \implies 12 + b = 0 \implies b = -12

The question asks for

a+b

, which is

-2 + (-12) = -14

. A: This is the value of

a

, not

a+b

. B: This is the value of

b

, not

a+b

. D: This answer may result from a sign error when solving for

b

Question 9

Evaluate the limit: $$ \lim_{x \to 2} \frac{\sqrt{x+7} - 3}{x-5}

0 (correct answer)
$\frac{1}{6}$
Does not exist
-\frac{1}{3}

Explanation: This question is a test of whether students check for indeterminate forms before applying more complex methods. The first step for any limit is direct substitution. Substitute

x = 2

into the expression: Numerator:

\sqrt{2+7} - 3 = \sqrt{9} - 3 = 3 - 3 = 0

. Denominator:

2 - 5 = -3

. The limit is

\frac{0}{-3} = 0

. B: This is the answer to a different, very common limit problem,

\lim_{x \to 2} \frac{\sqrt{x+7}-3}{x-2}

. Students who see the radical and assume they must use the conjugate method without first substituting may mistakenly arrive at this answer. C: This might be chosen if a student makes an error during substitution and believes the denominator is zero. D: This is the value of the denominator, not the limit itself.

Question 10

Evaluate the limit: $$ \lim_{x \to 1} \frac{x - 1}{x^{1/3} - 1}

$\frac{1}{3}$
0
3 (correct answer)
1

Explanation: This limit is of the form

\frac{0}{0}

. A helpful technique is to make a substitution. Let

u = x^{1/3}

. As

x \to 1

u \to 1^{1/3} = 1

. Also,

x = u^3

. Substituting these into the limit gives:

\lim_{u \to 1} \frac{u^3 - 1}{u - 1}

This is a standard limit that can be solved by factoring the numerator as a difference of cubes:

\lim_{u \to 1} \frac{(u-1)(u^2+u+1)}{u-1}

Cancel the

(u-1)

terms:

\lim_{u \to 1} (u^2+u+1) = 1^2 + 1 + 1 = 3

A: This is the reciprocal of the correct answer, which could result from inverting the expression. B: A common incorrect guess for

\frac{0}{0}

limits. D: This might result from an algebraic error after cancellation.

Question 11

Let $f(x)$ be a function such that for all $x \ne 1$ in an open interval containing 1, the inequality $1 - (x-1)^2 \le f(x) \le \cos(\pi(x-1))$ holds. What is $\lim_{x \to 1} f(x)$ ?

0
1 (correct answer)
Does not exist
-1

Explanation: This problem requires the use of the Squeeze Theorem. We must evaluate the limits of the lower and upper bounding functions as

x \to 1

. Limit of the lower bound:

\lim_{x \to 1} (1 - (x-1)^2) = 1 - (1-1)^2 = 1 - 0 = 1

Limit of the upper bound:

\lim_{x \to 1} \cos(\pi(x-1)) = \cos(\pi(1-1)) = \cos(0) = 1

Since

f(x)

is squeezed between two functions that both approach 1 as

x \to 1

, by the Squeeze Theorem,

\lim_{x \to 1} f(x)

must also be 1. A: This might be chosen if the first limit is incorrectly evaluated to 0. C: A student might choose this if they are unable to evaluate one of the limits. D: This is the value of

\cos(\pi)

, which would result from incorrectly evaluating the argument of cosine as

\pi

instead of 0.

Question 12

For what value of the constant $k$ does the following limit exist? $\lim_{x \to -1} \frac{x^2 - kx - 5}{x+1}$

-6
-4
4 (correct answer)
5

Explanation: For a limit of a rational function to exist where the denominator approaches 0, the numerator must also approach 0. This creates a

\frac{0}{0}

indeterminate form which may resolve to a finite value. Set the numerator equal to 0 at

x = -1

(-1)^2 - k(-1) - 5 = 0

1 + k - 5 = 0

k - 4 = 0

k = 4

A: This is the value of the limit itself if

k=4

\lim_{x \to -1} \frac{x^2 - 4x - 5}{x+1} = \lim_{x \to -1} \frac{(x-5)(x+1)}{x+1} = -1-5 = -6

. Students might solve for

k

and then solve for the limit, choosing the limit's value instead of

k

's. B: This results from a sign error in solving for

k

, such as

1-k-5=0

. D: This may result from misplacing a term in the setup equation.

Question 13

Evaluate the limit: $\lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4}$

$-3$ (correct answer)
$-1$
$3$
Does Not Exist

Explanation: Direct substitution of

x = -2

results in the indeterminate form

0/0

. To evaluate the limit, we must factor the numerator and the denominator. The numerator is a sum of cubes,

x^3 + 2^3 = (x+2)(x^2 - 2x + 4)

. The denominator is a difference of squares,

x^2 - 4 = (x-2)(x+2)

. The expression becomes

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

. We can cancel the

(x+2)

term, leaving

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

. Now, substituting

x = -2

gives

\frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = \frac{12}{-4} = -3

Question 14

Let $f$ be a function such that $-x^2 + 6x - 7 \le f(x) \le x^2 - 4x + 5$ for all $x$ . Find $\lim_{x \to 2} (x \cdot f(x))$ .

$0$
$1$
Does Not Exist
$2$ (correct answer)

Explanation: When you encounter a limit problem involving inequalities that bound a function, think about the Squeeze Theorem (also called the Sandwich Theorem). This powerful tool lets you find limits even when you don't know the exact form of the function. First, let's examine what happens to the bounding functions as

x

approaches 2. For the lower bound:

-x^2 + 6x - 7

, when

x = 2

, we get

-4 + 12 - 7 = 1

. For the upper bound:

x^2 - 4x + 5

, when

x = 2

, we get

4 - 8 + 5 = 1

. Since both bounds approach the same value,

\lim_{x \to 2} f(x) = 1

by the Squeeze Theorem. Now we can find

\lim_{x \to 2} (x \cdot f(x))

using the product rule for limits:

\lim_{x \to 2} x \cdot \lim_{x \to 2} f(x) = 2 \cdot 1 = 2

. Looking at the wrong answers: (A) 0 might tempt you if you incorrectly calculated one of the bounds or forgot to multiply by the limit of

x

. (B) 1 is a common error where you find

\lim_{x \to 2} f(x) = 1

correctly but forget to multiply by 2. (C) Does Not Exist would only be correct if the bounds approached different values, which they don't. Study tip: When you see a function squeezed between two expressions, immediately check if those bounds have the same limit at the point of interest. If they do, you can apply the Squeeze Theorem confidently, even without knowing the exact function.

Question 15

Evaluate the limit: $\lim_{x \to 1} \frac{\frac{1}{x+1} - \frac{1}{2}}{x-1}$

$-1/4$ (correct answer)
$1/4$
$1/2$
Does Not Exist

Explanation: Direct substitution of

x=1

results in the indeterminate form

0/0

. First, simplify the complex fraction in the numerator by finding a common denominator, which is

2(x+1)

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

We can factor out

-1

from the numerator:

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

. Now, cancel the

(x-1)

terms. The expression simplifies to

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

. The limit is now

\lim_{x \to 1} \frac{-1}{2(x+1)} = \frac{-1}{2(1+1)} = -\frac{1}{4}

Question 16

Suppose $\lim_{x \to a} f(x) = 4$ and $\lim_{x \to a} g(x) = -2$ . What is the value of $\lim_{x \to a} \frac{2f(x) - (g(x))^2}{\sqrt{f(x)}}$ ?

$5$
$2$ (correct answer)
$6$
$4$

Explanation: When you encounter limit problems with multiple functions, the key is applying limit laws systematically. These laws allow you to break down complex expressions into simpler parts when the individual limits exist. Since

\lim_{x \to a} f(x) = 4

and

\lim_{x \to a} g(x) = -2

, you can find the limit of the given expression by evaluating each component. First, find the numerator:

\lim_{x \to a} [2f(x) - (g(x))^2] = 2 \cdot 4 - (-2)^2 = 8 - 4 = 4

. Next, find the denominator:

\lim_{x \to a} \sqrt{f(x)} = \sqrt{4} = 2

. Therefore,

\lim_{x \to a} \frac{2f(x) - (g(x))^2}{\sqrt{f(x)}} = \frac{4}{2} = 2

. Let's examine why the other answers are incorrect. Choice A)

5

likely comes from miscalculating the numerator as

2(4) - (-2) = 10

instead of properly squaring

g(x)

. Choice C)

6

might result from errors like computing

2(4) - 2 = 6

(forgetting to square the negative) or making arithmetic mistakes in the final division. Choice D)

4

represents stopping at the numerator calculation without dividing by the denominator, essentially forgetting the fraction structure entirely. The correct answer is B)

2

. Remember this pattern: when working with limits of composite expressions, apply limit laws step-by-step to each operation (addition, multiplication, division, powers, roots). Always double-check your arithmetic, especially with negative numbers and exponents, as these are common sources of errors in limit calculations.

Question 17

Find the limit: $\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}$

$1/6$ (correct answer)
$1/3$
$0$
Does Not Exist

Explanation: Direct substitution of

h=0

yields the indeterminate form

0/0

. To resolve this, we multiply the numerator and denominator by the conjugate of the numerator, which is

\sqrt{9+h} + 3

. This gives:

\lim_{h \to 0} \frac{(\sqrt{9+h} - 3)(\sqrt{9+h} + 3)}{h(\sqrt{9+h} + 3)} = \lim_{h \to 0} \frac{(9+h) - 9}{h(\sqrt{9+h} + 3)} = \lim_{h \to 0} \frac{h}{h(\sqrt{9+h} + 3)}

After canceling

h

from the numerator and denominator, we have

\lim_{h \to 0} \frac{1}{\sqrt{9+h} + 3}

. Substituting

h=0

now yields

\frac{1}{\sqrt{9} + 3} = \frac{1}{3+3} = \frac{1}{6}

Question 18

Let $f$ and $g$ be functions such that $\lim_{x \to 1} g(x) = 8$ . If $f(x) = \sqrt[3]{x}$ is a continuous function, what is $\lim_{x \to 1} f(g(x))$ ?

$4$
$8$
$2$ (correct answer)
Cannot be determined

Explanation: When you encounter a composite function limit like this, you're dealing with the composition of limits theorem. This powerful tool tells you that if the inner function has a limit and the outer function is continuous at that limit value, then you can "move the limit inside" the outer function. Here's how to work through this step-by-step: You know that

\lim_{x \to 1} g(x) = 8

, and you're told that

f(x) = \sqrt[3]{x}

is continuous. Since

f

is continuous at

x = 8

(cube root is continuous everywhere), the composition of limits theorem applies:

\lim_{x \to 1} f(g(x)) = f\left(\lim_{x \to 1} g(x)\right) = f(8) = \sqrt[3]{8} = 2

. Let's examine why the other answers miss the mark. Choice A) gives 4, which might come from confusing this with

\sqrt{8} = 2\sqrt{2} \approx 2.83

or some other square root calculation. Choice B) gives 8, which represents the common error of thinking the limit equals

\lim_{x \to 1} g(x)

without applying the outer function

f

. Choice D) suggests the limit cannot be determined, but this is incorrect since we have all the information needed: the limit of the inner function and the continuity of the outer function. The key strategy here is recognizing when you can apply the composition of limits theorem. Whenever you see

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

, check if

g(x)

has a limit as

x \to a

and whether

f

is continuous at that limit value. If both conditions hold, you can evaluate the limit by substitution.

Question 19

Evaluate the limit: $$ \lim_{x \to -2} \frac{x^3 + 8}{x^2 - 4}

Does not exist
-3 (correct answer)
-1
6

Explanation: The limit is of the indeterminate form

\frac{0}{0}

. We can factor the numerator as a sum of cubes and the denominator as a difference of squares.

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

Cancel the

(x+2)

terms:

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

Now substitute

x = -2

\frac{(-2)^2 - 2(-2) + 4}{-2 - 2} = \frac{4 + 4 + 4}{-4} = \frac{12}{-4} = -3

A: This is a common guess for limits with a zero in the denominator before simplification. C: This results from an incorrect factorization of the sum of cubes, such as

(x+2)(x^2+2x+4)

. D: This results from sign errors during substitution, such as

\frac{4+4+4}{-2+2}

which is undefined, or

\frac{12}{2}

from sign errors in both numerator and denominator.

Question 20

Find the limit, which represents the derivative of a function $f(x) = x^{-2}$ at a point $x$ : $$ \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}

$-\frac{2}{x^3}$ (correct answer)
$\frac{2}{x^3}$
0
$-\frac{1}{x^2}$

Explanation: First, combine the fractions in the numerator by finding a common denominator:

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

Simplify the numerator:

\frac{-2xh - h^2}{h x^2(x+h)^2}

Factor out

h

from the numerator and cancel with the

h

in the denominator:

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

Now, evaluate the limit as

h \to 0

\lim_{h \to 0} \frac{-2x - h}{x^2(x+h)^2} = \frac{-2x - 0}{x^2(x+0)^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}

B: This choice results from a sign error when expanding

(x+h)^2

or simplifying the numerator. C: This results from an incorrect expansion of

(x+h)^2

x^2+h^2

, which leads to an incorrect simplification. D: This results from an error in algebraic simplification, possibly by incorrectly cancelling terms.