Calculus 1 Quiz: Continuity At A Point

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Continuity At A PointQuestion 1 of 20

The function $f(x) = \begin{cases} \frac{e^{2x}-1}{x}, & x \neq 0 \\ k, & x=0 \end{cases}$ is continuous at $x=0$ . What is the value of $k$ ?

k=0

k=1

k=2

k=e

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Calculus 1 Quiz

Calculus 1 Quiz: Continuity At A Point

Practice Continuity At A Point 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 Continuity At A Point, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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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

The function $f(x) = \begin{cases} \frac{e^{2x}-1}{x}, & x \neq 0 \\ k, & x=0 \end{cases}$ is continuous at $x=0$ . What is the value of $k$ ?

$k=0$
$k=1$
$k=2$ (correct answer)
$k=e$

Explanation: For

f

to be continuous at

x=0

k

must be equal to

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

. We need to evaluate

\lim_{x \to 0} \frac{e^{2x}-1}{x}

. This limit is of the indeterminate form

0/0

. We can use L'Hôpital's Rule.

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

Alternatively, this is the definition of the derivative of

g(x)=e^{2x}

x=0

g'(x)=2e^{2x}

, so

g'(0)=2

. Thus,

k=2

Question 2

The function $f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ 2-x & \text{if } x \text{ is irrational} \end{cases}$ is continuous at which of the following points?

$x=0$
$x=1$ (correct answer)
$x=2$
Nowhere

Explanation: For the function to be continuous at a point

c

, the limit must exist. This can only happen if the values of the two pieces of the function approach each other as

x

approaches

c

. Therefore, we must have

c = 2-c

. Solving this equation gives

2c = 2

, so

c = 1

. At

x=1

, both pieces of the function evaluate to 1. For any sequence of points approaching 1, rational or irrational, the function values will approach 1. So,

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

. Since

1

is rational,

f(1)=1

. The limit equals the function value, so

f

is continuous at

x=1

only.

Question 3

The function $f(x) = \frac{x^2+ax-10}{x-2}$ has a removable discontinuity at $x=2$ . What is the value of the constant $a$ ?

$a = -3$
$a = -1$
$a = 3$ (correct answer)
$a = 5$

Explanation: For

f(x)

to have a removable discontinuity at

x=2

, the limit

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

must exist. This requires the numerator to be zero at

x=2

, so that the factor

(x-2)

can be cancelled.

(2)^2 + a(2) - 10 = 0

4 + 2a - 10 = 0

2a - 6 = 0

a = 3

a=3

, then

f(x) = \frac{x^2+3x-10}{x-2} = \frac{(x-2)(x+5)}{x-2} = x+5

for

x \neq 2

. The limit as

x \to 2

is 7, so it is a removable discontinuity.

Question 4

Let the function $f$ be defined by $f(x) = \begin{cases} kx^2 + 2x & x < 2 \\ x^3 - kx & x \ge 2 \end{cases}$ . For what value of the constant $k$ is the function $f$ continuous at $x=2$ ?

$k = \frac{1}{2}$
$k = \frac{2}{3}$ (correct answer)
$k = 1$
$k = \frac{4}{3}$

Explanation: For

f

to be continuous at

x=2

, the left-hand limit must equal the right-hand limit, and this must equal the function value. We set

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

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

k(2)^2 + 2(2) = (2)^3 - k(2)

4k + 4 = 8 - 2k

6k = 4

k = \frac{4}{6} = \frac{2}{3}

With

k=\frac{2}{3}

, the limit is

\frac{20}{3}

and

f(2)=\frac{20}{3}

, so all conditions for continuity are met.

Question 5

How should $f(0)$ be defined to make the function $f(x) = \frac{(1+x)^3 - 1}{x}$ continuous at $x=0$ ?

$f(0) = 0$
$f(0) = 1$
$f(0) = 2$
$f(0) = 3$ (correct answer)

Explanation: For

f

to be continuous at

x=0

, we must define

f(0)

to be equal to the limit of

f(x)

x

approaches 0. We can find this limit by expanding the numerator:

\lim_{x \to 0} \frac{(1+x)^3 - 1}{x} = \lim_{x \to 0} \frac{(1 + 3x + 3x^2 + x^3) - 1}{x}

= \lim_{x \to 0} \frac{3x + 3x^2 + x^3}{x}

= \lim_{x \to 0} (3 + 3x + x^2)

= 3 + 3(0) + (0)^2 = 3

Therefore, we should define

f(0) = 3

Question 6

Let $f(x) = \begin{cases} \frac{x^2-(c+1)x+c}{x-c} & x \neq c \\ 5 & x = c \end{cases}$ . For which value of $c$ is $f(x)$ continuous for all $x$ ?

$c = 4$
$c = 5$
$c = 6$ (correct answer)
There is no such value of $c$ .

Explanation: For

f

to be continuous at

x=c

, we need

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

. Let's evaluate the limit. We can factor the numerator:

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

\lim_{x \to c} \frac{(x-c)(x-1)}{x-c} = \lim_{x \to c} (x-1) = c-1

Now, we set this limit equal to the function's value at

c

c-1 = f(c) = 5

c = 6

Thus, the function is continuous for all

x

c=6

Question 7

Let $a$ be a constant and let the function $f$ be defined as $f(x) = \frac{x^2 - (a+1)x + a}{x-a}$ . In order for $f$ to be continuous at $x=a$ , how must $f(a)$ be defined?

$f(a) = a$
$f(a) = 0$
$f(a) = 1$
$f(a) = a-1$ (correct answer)

Explanation: For

f

to be continuous at

x=a

, the value of

f(a)

must be equal to the limit of

f(x)

x

approaches

a

. The expression for

f(x)

gives an indeterminate form

\frac{0}{0}

when

x=a

, which suggests we can simplify the expression by factoring the numerator. The numerator is

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

. We can factor by grouping:

x(x-a) - 1(x-a) = (x-a)(x-1)

. So, for

x \neq a

, the function is

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

. Now we can find the limit:

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

. For continuity, we must define

f(a)

to be equal to this limit. Therefore,

f(a)

must be defined as

a-1

Question 8

Let $f$ be the function defined by the piecewise expression below.

For what value of the constant $k$ is the function $f$ continuous at $x=3$ ?

$k = \frac{1}{3}$
$k = \frac{5}{3}$
$k = \frac{11}{3}$ (correct answer)
$k = 6$

Explanation: For

f

to be continuous at

x=3

, three conditions must be met:

f(3)

must be defined, the limit

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

must exist, and

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

Find the limit from the left: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3^-} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3^-} (x+3) = 6$ .
Find the limit from the right: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (kx - 5) = 3k - 5$ .
For the overall limit to exist, the left and right limits must be equal: $6 = 3k - 5$ .
The value of the function at $x=3$ is $f(3) = k(3) - 5 = 3k - 5$ . For continuity, this must also equal the limit. So we solve $6 = 3k - 5$ , which gives $11 = 3k$ , so $k = \frac{11}{3}$ .

Question 9

Suppose for a function $f$ and a real number $c$ , it is known that $\lim_{x \to c} f(x) = L$ , where $L$ is a finite real number. Which of the following additional conditions is both necessary and sufficient to guarantee that $f$ is continuous at $x=c$ ?

The function $f$ must be defined at $x=c$ .
The value of $f(c)$ must be equal to $L$ . (correct answer)
The function $f$ must be differentiable at $x=c$ .
The value of $f(c)$ must be equal to $0$ .

Explanation: The definition of continuity at a point

c

requires three conditions: (1)

f(c)

is defined, (2)

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

exists, and (3)

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

. The stem states that condition (2) is met, with the limit being

L

. Choice A, $f(c)$ is defined, is necessary for continuity, but not sufficient. For example,

f(c)

could be defined as

L+1

, in which case condition (3) would fail. Choice B, $f(c)=L$ , combines conditions (1) and (3). If

f(c)=L

, then

f(c)

is defined and it equals the limit. This is precisely the third condition required for continuity and is therefore both necessary and sufficient, given that the limit exists. Choice C, differentiability at

c

, is a sufficient condition for continuity, but it is not necessary. A function can be continuous without being differentiable (e.g.,

f(x)=|x|

x=0

). Choice D is incorrect; there is no general requirement that

f(c)

must be 0 for continuity.

Question 10

Consider the function $f(x) = \frac{x^2 - 4}{|x - 2|}$ . Which of the following statements best describes the function $f$ at the point $x=2$ ?

The function has a removable discontinuity because the limit as $x \to 2$ exists.
The function has an infinite discontinuity because the denominator is zero at $x=2$ .
The function has a jump discontinuity because the one-sided limits at $x=2$ exist but are not equal. (correct answer)
The function is continuous at $x=2$ .

Explanation: To analyze the continuity at

x=2

, we must investigate the one-sided limits. The absolute value function

|x-2|

behaves differently depending on whether

x

is greater or less than 2. For the right-hand limit (

x \to 2^+

x > 2

, so

|x-2| = x-2

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

. For the left-hand limit (

x \to 2^-

x < 2

, so

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

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

. Since the left-hand limit (

-4

) and the right-hand limit (

4

) exist but are not equal, the overall limit

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

does not exist. This defines a jump discontinuity.

Question 11

Let $f(x) = x-2$ and let $g(x)$ be a piecewise function defined as $g(x) = \begin{cases} -1 & \text{if } x < 2 \\ 1 & \text{if } x \ge 2 \end{cases}$ . If $h(x) = f(x)g(x)$ , which of the following statements is true about the continuity of $h(x)$ at $x=2$ ?

$h(x)$ is not continuous at $x=2$ because $g(x)$ is not continuous at $x=2$ .
$h(x)$ is continuous at $x=2$ . (correct answer)
$h(x)$ has a removable discontinuity at $x=2$ .
$h(x)$ has a jump discontinuity at $x=2$ .

Explanation: To determine if

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

is continuous at

x=2

, we check the three conditions for continuity.

Check if $h(2)$ is defined: $h(2) = f(2)g(2) = (2-2)(1) = 0$ . It is defined.
Check if $\lim_{x \to 2} h(x)$ exists. We evaluate the one-sided limits: Left-hand limit: $\lim_{x \to 2^-} h(x) = \lim_{x \to 2^-} (x-2)g(x) = \lim_{x \to 2^-} (x-2)(-1) = (2-2)(-1) = 0$ . Right-hand limit: $\lim_{x \to 2^+} h(x) = \lim_{x \to 2^+} (x-2)g(x) = \lim_{x \to 2^+} (x-2)(1) = (2-2)(1) = 0$ . Since the left-hand and right-hand limits are equal, the limit exists and $\lim_{x \to 2} h(x) = 0$ .
Check if $\lim_{x \to 2} h(x) = h(2)$ . We found that both are equal to 0. Since all three conditions are satisfied, $h(x)$ is continuous at $x=2$ . The discontinuity of $g(x)$ is 'cancelled out' because $f(x)$ approaches 0 at $x=2$ .

Question 12

Let $f$ and $g$ be functions. If it is known that $g$ is continuous at $x=c$ and $f$ is continuous at $x=g(c)$ , which of the following statements about the composite function $h(x) = f(g(x))$ must be true?

$h(x)$ is continuous at $x=c$ . (correct answer)
$h(x)$ is continuous at $x=c$ only if $f$ is also continuous at $x=c$ .
$h(x)$ might have a removable discontinuity at $x=c$ .
$h(x)$ might have a jump discontinuity at $x=c$ .

Explanation: This question tests the theorem on the continuity of composite functions. The theorem states that if

g

is continuous at

c

and

f

is continuous at

g(c)

, then the composite function

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

is continuous at

c

. The conditions given in the stem are precisely the hypotheses of this theorem. To prove this from the definition of continuity:

Since $g$ is continuous at $c$ , $\lim_{x \to c} g(x) = g(c)$ .
Since $f$ is continuous at $g(c)$ , we know that $\lim_{y \to g(c)} f(y) = f(g(c))$ .
Combining these, we can evaluate the limit of the composite function: $\lim_{x \to c} h(x) = \lim_{x \to c} f(g(x))$ . Since $g(x) \to g(c)$ as $x \to c$ , we can substitute to get $\lim_{y \to g(c)} f(y)$ , which is $f(g(c))$ .
The value of the function at $c$ is $h(c) = f(g(c))$ . Since $\lim_{x \to c} h(x) = h(c)$ , the function $h$ is continuous at $c$ . The other choices are incorrect because the conditions provided are sufficient to guarantee continuity.

Question 13

Let $f$ be a function such that $f(1) = 2$ and $f(3) = 5$ . A student concludes that there must be a number $c$ in the interval $(1, 3)$ such that $f(c) = 4$ . This conclusion is guaranteed to be correct if which of the following conditions is also known to be true?

$f$ is defined on the interval $[1, 3]$ .
$f$ is increasing on the interval $[1, 3]$ .
$f$ is continuous on the interval $[1, 3]$ . (correct answer)
$f$ is differentiable on the interval $(1, 3)$ .

Explanation: The student's conclusion is an application of the Intermediate Value Theorem (IVT). The IVT states that if a function

f

is continuous on a closed interval

[a, b]

, and

N

is any number between

f(a)

and

f(b)

, then there exists at least one number

c

(a, b)

such that

f(c) = N

. In this case,

a=1

b=3

f(1)=2

f(3)=5

, and

N=4

. Since 4 is between 2 and 5, the theorem guarantees a value

c

(1,3)

with

f(c)=4

provided its main hypothesis is met. The crucial hypothesis for the IVT is that the function must be continuous on the closed interval

[1, 3]

. Choice A is not sufficient; a function can be defined but have a jump discontinuity that 'jumps over' the value 4. Choice B is not sufficient; an increasing function can still have a jump discontinuity. Choice D is not sufficient because differentiability on the open interval

(1, 3)

does not guarantee continuity on the closed interval

[1, 3]

(for example, at the endpoints). Continuity on the closed interval is the specific condition required for the IVT.

Question 14

Let the function $f$ be defined as $f(x) = \begin{cases} x^k \cos(\frac{\pi}{x}) & \text{if } x \neq 0 \\ L & \text{if } x = 0 \end{cases}$ . If $f$ is continuous at $x=0$ , what must be true about the constants $k$ and $L$ ?

$k > 0$ and $L = 0$ (correct answer)
$k < 0$ and $L = 0$
$k$ can be any real number and $L = 0$
$k > 0$ and $L$ can be any real number

Explanation: For

f

to be continuous at

x=0

, we must have

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

. Here,

f(0) = L

. So we need to evaluate

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

. We can use the Squeeze Theorem. The cosine function is bounded:

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

for all

x \neq 0

. Multiplying by

x^k

(assuming

x^k > 0

, we can use

|x^k|

for the general case):

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

. For the Squeeze Theorem to imply that the limit is 0, we need the limits of the outer functions to be 0. That is, we need

\lim_{x \to 0} |x^k| = 0

. This condition holds if and only if

k > 0

. If

k>0

, then

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

. For continuity, we must have

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

, so

L=0

. Therefore, we need

k>0

and

L=0

Question 15

Let $f(x) = \frac{1}{x-2}$ . The Intermediate Value Theorem states that for a continuous function on $[a, b]$ , the function takes on every value between $f(a)$ and $f(b)$ . Since $f(1)=-1$ and $f(3)=1$ , why does the theorem not guarantee a root $c$ in $(1,3)$ where $f(c)=0$ ?

The function does not take on the value 0.
The interval $(1,3)$ is open, but the theorem requires a closed interval.
The values $f(1)$ and $f(3)$ are not equal.
The function is not continuous on the interval $[1,3]$ . (correct answer)

Explanation: The Intermediate Value Theorem has a crucial precondition: the function must be continuous on the closed interval

[a, b]

. The function

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

has an infinite discontinuity at

x=2

, which is within the interval

[1,3]

. Because the function is not continuous on this interval, the theorem's conclusion is not guaranteed.

Question 16

Let $f(x) = x - \lfloor x \rfloor$ , where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ . Which statement best describes the continuity of $f$ at $x=3$ ?

$f$ has a jump discontinuity at $x=3$ because the one-sided limits are not equal. (correct answer)
$f$ has a removable discontinuity at $x=3$ because $f(3)$ can be redefined.
$f$ is continuous at $x=3$ because $f(3)=0$ and the right-hand limit is 0.
$f$ has an infinite discontinuity at $x=3$ .

Explanation: We test the three conditions for continuity at

x=3

Function value: $f(3) = 3 - \lfloor 3 \rfloor = 3 - 3 = 0$ . The function is defined at $x=3$ .
Left-hand limit: As $x$ approaches 3 from the left, $x$ is slightly less than 3 (e.g., 2.999). For such values, $\lfloor x \rfloor = 2$ . So, $\lim_{x \to 3^-} (x - \lfloor x \rfloor) = \lim_{x \to 3^-} (x - 2) = 3 - 2 = 1$ .
Right-hand limit: As $x$ approaches 3 from the right, $x$ is slightly greater than 3 (e.g., 3.001). For such values, $\lfloor x \rfloor = 3$ . So, $\lim_{x \to 3^+} (x - \lfloor x \rfloor) = \lim_{x \to 3^+} (x - 3) = 3 - 3 = 0$ . Since the left-hand limit (1) is not equal to the right-hand limit (0), the overall limit $\lim_{x \to 3} f(x)$ does not exist. This is the definition of a jump discontinuity.

Question 17

Let $f(x) = \frac{x^2 + 2x - 8}{x^3 - 8}$ . The function $f$ is discontinuous at $x=2$ . Which of the following actions would make $f$ continuous at $x=2$ ?

Define $f(2) = 0$ .
Define $f(2) = \frac{1}{2}$ . (correct answer)
Define $f(2) = \frac{3}{4}$ .
The function cannot be made continuous at $x=2$ because it has a non-removable discontinuity.

Explanation: A function has a removable discontinuity at a point

c

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

exists but is not equal to

f(c)

(or

f(c)

is undefined). To make the function continuous, we must define

f(c)

to be equal to the limit. First, find the limit of

f(x)

x \to 2

. We factor the numerator and denominator: Numerator:

x^2 + 2x - 8 = (x+4)(x-2)

. Denominator (difference of cubes):

x^3 - 8 = (x-2)(x^2 + 2x + 4)

. So, for

x \neq 2

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

. Now we can evaluate the limit:

\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x+4}{x^2 + 2x + 4} = \frac{2+4}{2^2 + 2(2) + 4} = \frac{6}{4+4+4} = \frac{6}{12} = \frac{1}{2}

. Since the limit exists, the discontinuity is removable. To make

f

continuous at

x=2

, we must define

f(2)

to be the value of the limit. Thus, we should define

f(2) = \frac{1}{2}

Question 18

Let $f(x) = \begin{cases} \frac{1-\cos(x)}{x^2} & x \neq 0 \\ c & x=0 \end{cases}$ . For what value of $c$ is $f(x)$ continuous at $x=0$ ?

$c=0$
$c=\frac{1}{2}$ (correct answer)
$c=1$
No such value exists.

Explanation: For continuity at

x=0

, we need

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

. This limit has the indeterminate form

\frac{0}{0}

, so we can apply L'Hôpital's Rule twice. First application:

\lim_{x \to 0} \frac{\sin(x)}{2x}

(still

\frac{0}{0}

) Second application:

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

Therefore,

c = \frac{1}{2}

Question 19

The function $f(x) = \begin{cases} \frac{\tan(kx)}{x} & x \neq 0 \\ 4 & x=0 \end{cases}$ is continuous at $x=0$ . What is the value of $k$ ?

$k=1$
$k=2$
$k=4$ (correct answer)
$k=\frac{1}{4}$

Explanation: For continuity at

x=0

, we need

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

. We are given

f(0)=4

. We evaluate the limit:

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

= \lim_{x \to 0} \left( \frac{\sin(kx)}{kx} \cdot k \cdot \frac{1}{\cos(kx)} \right)

x \to 0

\frac{\sin(kx)}{kx} \to 1

and

\cos(kx) \to \cos(0) = 1

. So the limit is:

(1) \cdot k \cdot \frac{1}{1} = k

Setting the limit equal to the function value:

k=4

Question 20

Let $f(x) = \begin{cases} a\cos(x) & x \le 0 \\ x^2+2a-3 & x > 0 \end{cases}$ . For what value of $a$ is $f$ continuous at $x=0$ ?

$a = -3$
$a = 1$
$a = 3$ (correct answer)
No such value of $a$ exists.

Explanation: For continuity at

x=0

, we must have

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

. Let's calculate the limits and the function value.

f(0) = a\cos(0) = a \cdot 1 = a

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} a\cos(x) = a\cos(0) = a

\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2+2a-3) = 0^2+2a-3 = 2a-3

. Equating the left and right limits:

a = 2a-3

. Solving for

a

gives

-a = -3

, so

a=3