Calculus 1 Quiz: Types Of Discontinuities

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Types Of DiscontinuitiesQuestion 1 of 20

Let $f(x) = x \cot(x)$ . Which of the following describes a discontinuity of $f(x)$ ?

A removable discontinuity at

x=0

An infinite discontinuity at

x=0

A removable discontinuity at

x=\pi/2

An infinite discontinuity at

x=\pi/2

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Calculus 1 Quiz

Calculus 1 Quiz: Types Of Discontinuities

Practice Types Of Discontinuities 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 Types Of Discontinuities, 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

Let $f(x) = x \cot(x)$ . Which of the following describes a discontinuity of $f(x)$ ?

A removable discontinuity at $x=0$ . (correct answer)
An infinite discontinuity at $x=0$ .
A removable discontinuity at $x=\pi/2$ .
An infinite discontinuity at $x=\pi/2$ .

Explanation: The function can be written as

f(x) = \frac{x \cos(x)}{\sin(x)}

. Discontinuities occur when

\sin(x)=0

, which is at

x = n\pi

for any integer

n

. Let's analyze

x=0

: We need to find the limit

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

. We can rewrite this as

\lim_{x \to 0} \frac{x}{\sin(x)} \cdot \cos(x)

. We know that

\lim_{x \to 0} \frac{\sin(x)}{x} = 1

, so

\lim_{x \to 0} \frac{x}{\sin(x)} = 1

. Also,

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

. Therefore, the overall limit is

1 \cdot 1 = 1

. Since the limit exists but

f(0)

is undefined, the discontinuity at

x=0

is removable. Let's analyze

x=\pi/2

: At this point,

f(\pi/2) = \frac{(\pi/2)\cos(\pi/2)}{\sin(\pi/2)} = \frac{(\pi/2) \cdot 0}{1} = 0

. The function is defined and continuous at

x=\pi/2

. Let's analyze

x=\pi

: The limit is

\lim_{x \to \pi} \frac{x \cos(x)}{\sin(x)}

. The numerator approaches

\pi \cos(\pi) = -\pi

. The denominator approaches

\sin(\pi) = 0

. This is an infinite discontinuity.

Question 2

Let $f(x) = \frac{1-e^{x-2}}{x^2-4}$ . Which of the following describes the type of discontinuity at $x=2$ ?

Infinite discontinuity
Jump discontinuity
Removable discontinuity (correct answer)
Oscillating discontinuity

Explanation: The function has a potential discontinuity at

x=2

because the denominator is zero. To classify it, we must evaluate the limit

\lim_{x \to 2} \frac{1-e^{x-2}}{x^2-4}

. This is an indeterminate form of type

0/0

. We can use L'Hôpital's Rule. The derivative of the numerator is

\frac{d}{dx}(1-e^{x-2}) = -e^{x-2}

. The derivative of the denominator is

\frac{d}{dx}(x^2-4) = 2x

. Applying L'Hôpital's Rule:

\lim_{x \to 2} \frac{-e^{x-2}}{2x} = \frac{-e^{2-2}}{2(2)} = \frac{-e^0}{4} = -\frac{1}{4}

. Since the limit exists and is finite, but the function is undefined at

x=2

, the discontinuity is removable.

Question 3

Let the function $f$ be defined as follows: $f(x) = \begin{cases} \frac{x^2 - 9}{x-3} & \text{if } x < 3 \\ k & \text{if } x = 3 \\ x^2 - 3x + c & \text{if } x > 3 \end{cases}$ If $f(x)$ has a removable discontinuity at $x=3$ , which of the following statements must be true?

$c=6$ and $k=6$
$c=6$ and $k \neq 6$ (correct answer)
$c \neq 6$ and $k=6$
$c \neq 6$ and $k \neq 6$

Explanation: For a discontinuity at

x=3

to be removable, the limit

\lim_{x\to 3} f(x)

must exist. This requires the left-hand limit and the right-hand limit to be equal.
Left-hand limit:

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

.
Right-hand limit:

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

.
For the limit to exist, we must have

c=6

.
For there to be a discontinuity, the limit must not be equal to the function's value at that point, i.e.,

\lim_{x\to 3} f(x) \neq f(3)

.
Since

\lim_{x\to 3} f(x) = 6

and

f(3) = k

, we must have

k \neq 6

.
Therefore, we must have

c=6

and

k \neq 6

Question 4

Consider the function $f(x) = x - \lfloor x \rfloor$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ . What type of discontinuity does $f(x)$ have at each integer $x=n$ ?

A jump discontinuity (correct answer)
A removable discontinuity
An infinite discontinuity
The function is continuous at each integer

Explanation: Let's analyze the function at an arbitrary integer

x=n

. We need to check the one-sided limits.
Left-hand limit: As

x \to n^-

x

is slightly less than

n

, so

\lfloor x \rfloor = n-1

. Therefore,

\lim_{x\to n^-} (x - \lfloor x \rfloor) = n - (n-1) = 1

.
Right-hand limit: As

x \to n^+

x

is slightly greater than

n

, so

\lfloor x \rfloor = n

. Therefore,

\lim_{x\to n^+} (x - \lfloor x \rfloor) = n - n = 0

.
Since the left-hand limit (1) and the right-hand limit (0) both exist but are unequal, the function has a jump discontinuity at every integer

n

Question 5

Suppose $f(x)$ has a jump discontinuity at $x=c$ and $g(x)$ is a function that is continuous at $x=c$ . Which of the following statements about the function $h(x) = f(x) + g(x)$ at $x=c$ is always true?

$h(x)$ has an infinite discontinuity.
$h(x)$ has a removable discontinuity.
$h(x)$ is continuous.
$h(x)$ has a jump discontinuity. (correct answer)

Explanation: When you encounter problems about combining functions with different types of discontinuities, the key principle is that discontinuities in one function will "carry over" to the combined function unless something special happens to cancel them out. Let's think about what happens when we add

f(x)

and

g(x)

x = c

. Since

f(x)

has a jump discontinuity at

x = c

, the left-hand and right-hand limits of

f(x)

exist but are unequal:

\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)

. Meanwhile,

g(x)

is continuous at

x = c

, so

\lim_{x \to c^-} g(x) = \lim_{x \to c^+} g(x) = g(c)

. For

h(x) = f(x) + g(x)

, we can use limit properties:

\lim_{x \to c^-} h(x) = \lim_{x \to c^-} f(x) + g(c)

and

\lim_{x \to c^+} h(x) = \lim_{x \to c^+} f(x) + g(c)

. Since the limits of

f(x)

from left and right differ, adding the same constant

g(c)

to both sides preserves this difference. Therefore,

h(x)

has a jump discontinuity at

x = c

. Choice A is wrong because infinite discontinuities involve limits approaching infinity, which doesn't happen here. Choice B is incorrect because removable discontinuities can be "fixed" by redefining the function at a point, but our jump can't be removed. Choice C is wrong because the jump from

f(x)

prevents

h(x)

from being continuous. Remember: when combining functions, discontinuities persist unless there's a specific mechanism to cancel them out. Adding a continuous function doesn't eliminate jumps.

Question 6

Consider the function $f(x) = \begin{cases} \frac{x-1}{|x-1|} & \text{if } x \neq 1 \\ 0 & \text{if } x = 1 \end{cases}$ . What is the nature of the discontinuity of the function $g(x) = xf(x)$ at $x=1$ ?

Removable
Jump (correct answer)
Infinite
Continuous

Explanation: First, let's analyze

f(x)

. For

x>1

f(x)=1

. For

x<1

f(x)=-1

. This is a jump discontinuity at

x=1

. Now we analyze

g(x)=xf(x)

x=1

. We check the one-sided limits:

\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} x \cdot (1) = 1

\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} x \cdot (-1) = -1

. Since the left-hand limit (-1) and the right-hand limit (1) exist but are not equal, the function

g(x)

has a jump discontinuity at

x=1

. The value

g(1)=1 \cdot f(1) = 1 \cdot 0 = 0

does not affect this classification.

Question 7

The function $f(x) = \frac{x^3 + x^2 - 12x}{x^2 - 9}$ has two discontinuities. Which of the following correctly classifies them?

Removable at $x=3$ , infinite at $x=-3$ . (correct answer)
Infinite at $x=3$ , removable at $x=-3$ .
Both are removable discontinuities.
Both are infinite discontinuities.

Explanation: First, factor the numerator and the denominator. Numerator:

x(x^2+x-12) = x(x+4)(x-3)

. Denominator:

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

. So,

f(x) = \frac{x(x+4)(x-3)}{(x-3)(x+3)}

. The discontinuities are at

x=3

and

x=-3

. At

x=3

: The factor

(x-3)

appears in both the numerator and the denominator. We can cancel it to find the limit:

\lim_{x \to 3} \frac{x(x+4)}{x+3} = \frac{3(3+4)}{3+3} = \frac{21}{6} = \frac{7}{2}

. Since the limit exists, the discontinuity at

x=3

is removable. At

x=-3

: The factor

(x+3)

is in the denominator but not the numerator (after cancellation). The limit is

\lim_{x \to -3} \frac{x(x+4)}{x+3}

. The numerator approaches

-3(-3+4) = -3

, while the denominator approaches 0. This is an infinite discontinuity.

Question 8

The function $f(x)$ has a removable discontinuity at $x=a$ and $g(x)$ has an infinite discontinuity at $x=a$ . Which of the following statements about the sum $h(x) = f(x) + g(x)$ at $x=a$ is always true?

$h(x)$ has a removable discontinuity.
$h(x)$ has a jump discontinuity.
$h(x)$ has an infinite discontinuity. (correct answer)
$h(x)$ could be continuous.

Explanation: A removable discontinuity for

f(x)

x=a

means that

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

for some finite number L. An infinite discontinuity for

g(x)

x=a

means that

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

. To analyze the sum

h(x) = f(x) + g(x)

, we look at its limit:

\lim_{x \to a} h(x) = \lim_{x \to a} (f(x) + g(x))

. Using the limit laws for sums, this is

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

. This becomes

L + (\pm\infty)

. Adding a finite number

L

to positive or negative infinity still results in positive or negative infinity. Therefore,

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

, which is the definition of an infinite discontinuity.

Question 9

A function $f(x)$ has a jump discontinuity at $x=c$ , and a function $g(x)$ is continuous for all real numbers. If the product $h(x) = f(x)g(x)$ is continuous at $x=c$ , which of the following MUST be true?

$g(c) = 0$ (correct answer)
$g(c) = 1$
$g(x)$ must be the constant function $g(x)=0$ .
The jump discontinuity in $f(x)$ cannot be removed by multiplication.

Explanation: Let the jump discontinuity at

x=c

for

f(x)

be characterized by

\lim_{x \to c^-} f(x) = L_1

and

\lim_{x \to c^+} f(x) = L_2

, where

L_1 \neq L_2

. Since

g(x)

is continuous,

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

. For

h(x) = f(x)g(x)

to be continuous at

x=c

, its left-hand and right-hand limits must be equal. Left-hand limit of

h(x)

\lim_{x \to c^-} f(x)g(x) = L_1 \cdot g(c)

. Right-hand limit of

h(x)

\lim_{x \to c^+} f(x)g(x) = L_2 \cdot g(c)

. Setting them equal gives

L_1 \cdot g(c) = L_2 \cdot g(c)

, which simplifies to

(L_1 - L_2) \cdot g(c) = 0

. Since

f(x)

has a jump discontinuity,

L_1 \neq L_2

, so

L_1 - L_2 \neq 0

. Therefore, we must have

g(c) = 0

Question 10

Describe the discontinuities of the function $f(x) = \frac{x^2 - x - 2}{x^3 - 2x^2}$ .

A removable discontinuity at $x=2$ and an infinite discontinuity at $x=0$ . (correct answer)
A jump discontinuity at $x=2$ and an infinite discontinuity at $x=0$ .
An infinite discontinuity at $x=2$ and a removable discontinuity at $x=0$ .
Infinite discontinuities at both $x=0$ and $x=2$ .

Explanation: First, factor the numerator and denominator:

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

. The function is undefined at

x=0

and

x=2

.
For

x \neq 2

, we can simplify the function to

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

.
To classify the discontinuity at

x=2

, we evaluate the limit:

\lim_{x\to 2} \frac{x+1}{x^2} = \frac{2+1}{2^2} = \frac{3}{4}

. Since the limit exists but

f(2)

is undefined, there is a removable discontinuity at

x=2

.
To classify the discontinuity at

x=0

, we evaluate the limit:

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

. As

x \to 0

, the numerator approaches 1 and the denominator

x^2

approaches 0 from the positive side. Thus, the limit is

+\infty

. This is an infinite discontinuity.

Question 11

Classify the discontinuity of the function $f(x) = \frac{x^2 - 5x + 6}{|x-3|}$ at $x=3$ .

Jump discontinuity (correct answer)
Removable discontinuity
Infinite discontinuity
Oscillating discontinuity

Explanation: To classify the discontinuity at

x=3

, we must analyze the left-hand and right-hand limits. First, factor the numerator:

f(x) = \frac{(x-2)(x-3)}{|x-3|}

.
For the right-hand limit (

x \to 3^+

x > 3

, so

|x-3| = x-3

\lim_{x\to 3^+} \frac{(x-2)(x-3)}{x-3} = \lim_{x\to 3^+} (x-2) = 3-2 = 1

.
For the left-hand limit (

x \to 3^-

x < 3

, so

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

\lim_{x\to 3^-} \frac{(x-2)(x-3)}{-(x-3)} = \lim_{x\to 3^-} -(x-2) = -(3-2) = -1

.
Since the left-hand limit (

-1

) and right-hand limit (

1

) both exist but are not equal, the function has a jump discontinuity at

x=3

Question 12

Consider the function $f(x) = \frac{x^2 - 1}{\sin(\pi x)}$ . Which of the following correctly describes the types of discontinuities for this function at integer values of $x$ ?

Removable at all integers where the function is discontinuous.
Removable at $x=1$ and $x=-1$ ; infinite at all other integers. (correct answer)
Infinite at all integers where the function is discontinuous.
Jump at $x=1$ and $x=-1$ ; infinite at all other integers.

Explanation: When analyzing discontinuities in rational functions, you need to identify where the function is undefined and then classify each discontinuity type. For

f(x) = \frac{x^2 - 1}{\sin(\pi x)}

, the function is undefined wherever

\sin(\pi x) = 0

, which occurs at all integer values of

x

. To classify each discontinuity, examine the behavior of both numerator and denominator at these points. At

x = 1

: the numerator equals

1^2 - 1 = 0

and the denominator equals

\sin(\pi) = 0

. This creates a

\frac{0}{0}

indeterminate form. Since

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

, you can factor out the

(x-1)

term, and the limit exists as

x

approaches 1. This makes the discontinuity removable. The same analysis applies at

x = -1

. At any other integer

n

(where

n \neq \pm 1

), the numerator

n^2 - 1 \neq 0

while the denominator still equals zero. This creates a

\frac{\text{nonzero}}{0}

situation, causing the function to approach

\pm\infty

, making these infinite discontinuities. Choice A is wrong because not all integer discontinuities are removable—only

x = \pm 1

. Choice C incorrectly claims all discontinuities are infinite, missing the removable ones at

x = \pm 1

. Choice D mentions jump discontinuities, but rational functions don't typically have jump discontinuities; they have removable or infinite types. Study tip: For rational functions, check if both numerator and denominator equal zero at the same point—that's your clue for a potentially removable discontinuity.

Question 13

Consider the function $f$ defined by $f(x) = \begin{cases} x^2 \cos(\frac{1}{x}) & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$ What type of discontinuity does $f(x)$ have at $x=0$ ?

A jump discontinuity
An oscillating discontinuity
A removable discontinuity (correct answer)
The function is continuous at $x=0$

Explanation: When analyzing discontinuities in piecewise functions, you need to examine what happens to the function's limit as you approach the point in question, then compare that to the function's actual value at that point. To determine the type of discontinuity at

x = 0

, let's find

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

. For

x \neq 0

, we have

f(x) = x^2 \cos(\frac{1}{x})

. Since

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

for all

x \neq 0

, we can use the squeeze theorem:

-x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2

x \to 0

, both

-x^2

and

x^2

approach 0, so

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

. However,

f(0) = 1 \neq 0

. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. Option A is incorrect because a jump discontinuity occurs when left and right limits exist but are unequal. Here, both one-sided limits equal 0. Option B is wrong because while

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

oscillates wildly near zero, the

x^2

factor forces the overall limit to exist—the oscillation doesn't prevent convergence. Option D is incorrect since

f(0) = 1

while

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

, creating a discontinuity. Remember: A discontinuity is removable when the limit exists but either the function isn't defined at that point or has the wrong value. You could "remove" the discontinuity by redefining

f(0) = 0

Question 14

Consider the function $f(x) = \frac{2 - \sqrt{x+3}}{x-1}$ . Which of the following statements correctly describes the discontinuity at $x=1$ ?

It is an infinite discontinuity.
It is a removable discontinuity, and the limit is $1/4$ .
It is a removable discontinuity, and the limit is $-1/4$ . (correct answer)
It is a jump discontinuity.

Explanation: When you encounter a rational function that's undefined at a particular point, you need to determine whether the discontinuity is removable, infinite, or a jump. The key is examining what happens to the function values as you approach the problematic point. For

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

, the function is undefined at

x = 1

because the denominator equals zero. To classify this discontinuity, let's find

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

. Direct substitution gives

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

, an indeterminate form. This suggests we might have a removable discontinuity if we can resolve this form. To eliminate the indeterminate form, multiply by the conjugate:

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

Since

1-x = -(x-1)

, we get:

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

Since this limit exists, the discontinuity is removable, making (C) correct. (A) is wrong because the limit exists and is finite, not infinite. (B) has the wrong limit value—a sign error from the rationalization process. (D) is incorrect because jump discontinuities occur when left and right limits exist but differ, which isn't the case here. Study tip: When you see

\frac{0}{0}

, try algebraic manipulation like factoring or rationalizing before concluding the limit doesn't exist.

Question 15

If a function $f(x)$ has a non-removable discontinuity at $x=c$ , which of the following statements must be true?

The one-sided limits, $\lim_{x\to c^-} f(x)$ and $\lim_{x\to c^+} f(x)$ , must exist but be unequal.
The limit $\lim_{x\to c} f(x)$ must be equal to $\infty$ or $-\infty$ , indicating an infinite discontinuity.
It is not possible to make the function continuous at $x=c$ by redefining the value of $f(c)$ . (correct answer)
The function $f(c)$ must be undefined, making it impossible to evaluate at that point.

Explanation: When you encounter questions about non-removable discontinuities, focus on what distinguishes them from removable discontinuities. The key insight is that non-removable discontinuities represent fundamental "breaks" in a function that cannot be fixed simply by changing or defining the function value at a single point. A non-removable discontinuity means the function has an inherent structural problem at

x = c

that prevents continuity, regardless of how you define

f(c)

. This is exactly what option C describes - you cannot make the function continuous at

x = c

by redefining

f(c)

, which is the defining characteristic of a non-removable discontinuity. Option A is incorrect because while jump discontinuities (where one-sided limits exist but differ) are one type of non-removable discontinuity, they're not the only type. Non-removable discontinuities also include cases where one or both one-sided limits don't exist. Option B is wrong because infinite discontinuities are just one specific type of non-removable discontinuity. Non-removable discontinuities also include jump discontinuities and oscillating discontinuities, where the limits aren't infinite. Option D is incorrect because whether

f(c)

is defined or undefined doesn't determine if a discontinuity is removable or non-removable. A function can have

f(c)

defined and still have a non-removable discontinuity if the limit doesn't exist or differs from

f(c)

in an unfixable way. Study tip: Remember that "non-removable" literally means "cannot be removed by redefining the function value" - this definition directly leads you to the correct answer.

Question 16

Identify the types of discontinuities for the function $f(x) = \frac{x^3 - 4x^2 + 4x}{x^2 - 4}$ .

Removable at both $x=2$ and $x=-2$ .
Infinite at $x=2$ , removable at $x=-2$ .
Infinite at both $x=2$ and $x=-2$ .
Removable at $x=2$ , infinite at $x=-2$ . (correct answer)

Explanation: When you encounter a rational function like this, you need to identify where discontinuities occur and classify their types. Discontinuities happen where the denominator equals zero, so start by factoring both numerator and denominator. First, factor the denominator:

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

, giving potential discontinuities at

x = 2

and

x = -2

. Next, factor the numerator:

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

So your function becomes:

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

for

x \neq 2

x = 2

: The factor

(x-2)

appears in both numerator and denominator, so it cancels out. This creates a removable discontinuity (a "hole") because you can define the limit. Using the simplified form,

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

x = -2

: The factor

(x+2)

appears only in the denominator after cancellation, so the function approaches infinity. This creates an infinite discontinuity (vertical asymptote). Now examine each choice: Choice A incorrectly claims both are removable. Choice B reverses the types of discontinuities. Choice C incorrectly claims both are infinite discontinuities. Choice D correctly identifies a removable discontinuity at

x = 2

and an infinite discontinuity at

x = -2

. Study tip: Always factor completely first, then check if common factors cancel (removable) or if denominators remain zero (infinite). The key is whether factors disappear from the denominator after simplification.

Question 17

The function $f(x) = \frac{3x^2 - 5x - 2}{x^2 - 4}$ has a removable discontinuity at $x=2$ . What value should be assigned to $f(2)$ to make the function continuous at $x=2$ ?

There is no value for which the function is continuous at $x=2$ .
$5/4$
$0$
$7/4$ (correct answer)

Explanation: When you encounter a rational function with a removable discontinuity, you're dealing with a hole in the graph that can be "filled" by defining the function value at that point. The key is recognizing that both the numerator and denominator equal zero at the discontinuous point, creating an indeterminate form that can be resolved through algebraic manipulation. Since

f(x) = \frac{3x^2 - 5x - 2}{x^2 - 4}

has a removable discontinuity at

x = 2

, both the numerator and denominator must equal zero when

x = 2

. Let's verify: the denominator

2^2 - 4 = 0

✓, and the numerator

3(4) - 5(2) - 2 = 12 - 10 - 2 = 0

✓. To find the correct value, factor both expressions. The denominator factors as

(x-2)(x+2)

. For the numerator, we need

3x^2 - 5x - 2 = (x-2)(3x+1)

. After canceling the common factor

(x-2)

\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{3x+1}{x+2} = \frac{3(2)+1}{2+2} = \frac{7}{4}

Choice A is wrong because the discontinuity is removable, not essential. Choice B (

\frac{5}{4}

) likely comes from incorrectly factoring the numerator. Choice C (

0

) might result from thinking the limit equals zero since both numerator and denominator are zero at

x = 2

. Remember: for removable discontinuities, always factor and simplify first, then evaluate the limit. The correct value to assign is always this limit value.

Question 18

Let $f(x) = \frac{|x-a|}{x^2 - a^2}$ for a non-zero constant $a$ . Which of the following correctly classifies the discontinuities of $f(x)$ ?

A removable discontinuity at $x=a$ and an infinite discontinuity at $x=-a$ .
A jump discontinuity at $x=a$ and an infinite discontinuity at $x=-a$ . (correct answer)
Infinite discontinuities at both $x=a$ and $x=-a$ .
A jump discontinuity at $x=a$ and a removable discontinuity at $x=-a$ .

Explanation: The function is

f(x) = \frac{|x-a|}{(x-a)(x+a)}

. We examine the points

x=a

and

x=-a

. At

x=a

: We analyze the one-sided limits. For

x > a

|x-a| = x-a

, so

f(x) = \frac{x-a}{(x-a)(x+a)} = \frac{1}{x+a}

. Thus,

\lim_{x \to a^+} f(x) = \frac{1}{2a}

. For

x < a

|x-a| = -(x-a)

, so

f(x) = \frac{-(x-a)}{(x-a)(x+a)} = \frac{-1}{x+a}

. Thus,

\lim_{x \to a^-} f(x) = \frac{-1}{2a}

. Since

a \neq 0

, the left and right limits exist but are not equal, which defines a jump discontinuity. At

x=-a

: The limit is

\lim_{x \to -a} \frac{|x-a|}{(x-a)(x+a)}

. The numerator approaches

|-2a| = 2|a|

, which is a positive number. The denominator approaches 0. This results in an infinite limit, which defines an infinite discontinuity.

Question 19

Let $f(x) = \frac{x^2 - 1}{|x|-1}$ . Which of the following statements accurately describes the discontinuities of this function?

A removable discontinuity at $x=1$ and a jump discontinuity at $x=-1$ .
Removable discontinuities at $x=1$ and $x=-1$ . (correct answer)
A jump discontinuity at $x=1$ and a removable discontinuity at $x=-1$ .
Jump discontinuities at $x=1$ and $x=-1$ .

Explanation: The function is undefined when

|x|-1 = 0

, which means at

x=1

and

x=-1

. Let's analyze these points. For

x>0

|x|=x

, so

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

(for

x\neq 1

). The limit as

x \to 1

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

. Since the limit exists but

f(1)

is undefined, there is a removable discontinuity at

x=1

. For

x<0

|x|=-x

, so

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

(for

x\neq -1

). The limit as

x \to -1

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

. Since the limit exists but

f(-1)

is undefined, there is also a removable discontinuity at

x=-1

Question 20

Consider the function $f(x) = \frac{\sin(2x-6)}{x^2-9}$ . Which of the following correctly describes the function's discontinuities?

An infinite discontinuity at $x=3$ and a removable discontinuity at $x=-3$ .
A removable discontinuity at $x=3$ and an infinite discontinuity at $x=-3$ . (correct answer)
Infinite discontinuities at both $x=3$ and $x=-3$ .
Removable discontinuities at both $x=3$ and $x=-3$ .

Explanation: The denominator

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

is zero at

x=3

and

x=-3

. At

x=3

: We evaluate the limit

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

. We can rewrite this as

\lim_{x \to 3} \frac{\sin(2(x-3))}{x-3} \cdot \frac{1}{x+3}

. Using the property

\lim_{u \to 0} \frac{\sin(ku)}{u} = k

, the first part of the limit is

\lim_{x \to 3} \frac{\sin(2(x-3))}{x-3} = 2

. The second part is

\lim_{x \to 3} \frac{1}{x+3} = \frac{1}{6}

. The overall limit is

2 \cdot \frac{1}{6} = \frac{1}{3}

. Since the limit exists, the discontinuity at

x=3

is removable. At

x=-3

: We evaluate the limit

\lim_{x \to -3} \frac{\sin(2x-6)}{x^2-9}

. The numerator approaches

\sin(-12)

, which is a non-zero constant. The denominator approaches 0. A non-zero numerator with a denominator approaching zero indicates an infinite discontinuity.