IB Mathematics: Analysis and Approaches Quiz: Limits And Continuity

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Limits And ContinuityQuestion 1 of 20

Let $f(x) = \frac{1}{x-1}$ and $g(x) = x^2$ . The composite function $h(x) = f(g(x))$ is discontinuous at which value(s) of $x$ ?

x=-1

and

x=1

x=0

and

x=1

x=1

only.

At no values of

x

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Limits And Continuity

Practice Limits And Continuity in IB Mathematics: Analysis and Approaches 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 And Continuity, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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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.

All questions

Question 1

Let $f(x) = \frac{1}{x-1}$ and $g(x) = x^2$ . The composite function $h(x) = f(g(x))$ is discontinuous at which value(s) of $x$ ?

At $x=-1$ and $x=1$ . (correct answer)
At $x=0$ and $x=1$ .
At $x=1$ only.
At no values of $x$ .

Explanation: When dealing with composite functions and discontinuities, you need to find where the outer function becomes undefined after substitution. A function is discontinuous where it's not defined or where there's a break in the graph. First, let's find the composite function

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

. Since

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

and

g(x) = x^2

, we substitute

g(x)

into

f(x)

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

Now,

h(x)

is discontinuous wherever the denominator equals zero. Setting

x^2 - 1 = 0

x^2 = 1

x = \pm 1

h(x)

is discontinuous at

x = -1

and

x = 1

. Looking at the answer choices: A correctly identifies both discontinuities at

x = -1

and

x = 1

. B incorrectly includes

x = 0

, which would make the denominator

0^2 - 1 = -1 \neq 0

, so the function is actually defined there. C only identifies

x = 1

, missing the discontinuity at

x = -1

. D claims there are no discontinuities, which ignores the fact that rational functions are undefined when their denominators equal zero. The correct answer is A. Study tip: For composite function discontinuities, always substitute completely first, then find where the result is undefined. Don't just look at where the individual functions are discontinuous—the composition can create new discontinuities or eliminate existing ones.

Question 2

Let $f(x) = \frac{1}{x-1}$ and $g(x) = x^2$ . The composite function $h(x) = f(g(x))$ is discontinuous at which value(s) of $x$ ?

At $x=-1$ and $x=1$ . (correct answer)
At $x=0$ and $x=1$ .
At $x=1$ only.
At no values of $x$ .

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

. Since

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

and

g(x) = x^2

, we substitute

g(x)

into

f(x)

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

Now,

h(x)

is discontinuous wherever the denominator equals zero. Setting

x^2 - 1 = 0

x^2 = 1

x = \pm 1

h(x)

is discontinuous at

x = -1

and

x = 1

. Looking at the answer choices: A correctly identifies both discontinuities at

x = -1

and

x = 1

. B incorrectly includes

x = 0

, which would make the denominator

0^2 - 1 = -1 \neq 0

, so the function is actually defined there. C only identifies

x = 1

, missing the discontinuity at

x = -1

Question 3

For which value of $a$ is the function $f(x) = \begin{cases} ae^x + 1 & \text{if } x \le 0 \\ x^2 + 3 & \text{if } x > 0 \end{cases}$ continuous at $x=0$ ?

1
2 (correct answer)
3
4

Explanation: For continuity at

x=0

, the left-hand limit must equal the right-hand limit. We calculate both limits. \ Left-hand limit:

\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (ae^x + 1) = a e^0 + 1 = a(1) + 1 = a+1

. \ Right-hand limit:

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

. \ Set the limits equal to each other:

a+1 = 3

. Solving for

a

gives

a=2

Question 4

Find the value of the limit $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$ .

-2
0
5 (correct answer)
Does not exist

Explanation: To evaluate this limit, we first try to substitute

x=3

, which results in the indeterminate form

\frac{0}{0}

. This suggests that

(x-3)

is a factor of the numerator. Factoring the numerator gives

x^2 - x - 6 = (x-3)(x+2)

. The expression becomes

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

. For

x \neq 3

, we can cancel the

(x-3)

terms, leaving

\lim_{x \to 3} (x+2)

. Now we can substitute

x=3

to get

3+2 = 5

Question 5

The function $f(x)$ is continuous on the interval $[-2, 5]$ . It is known that $f(-2)=3$ and $f(5)=-1$ .

Which theorem guarantees that there is a value $c$ in the interval $(-2, 5)$ such that $f(c)=0$ ?

The Squeeze Theorem
The Extreme Value Theorem
The Mean Value Theorem
The Intermediate Value Theorem (correct answer)

Explanation: The Intermediate Value Theorem (IVT) states that if a function

f

is continuous on a closed interval

[a, b]

, then for any value

N

between

f(a)

and

f(b)

, there exists at least one number

c

(a, b)

such that

f(c) = N

. In this case, the function is continuous on

[-2, 5]

f(-2)=3

, and

f(5)=-1

. Since 0 is a value between -1 and 3, the IVT guarantees that there is a value

c

(-2, 5)

such that

f(c)=0

Question 6

Consider the piecewise function defined by $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 2x + 1 & \text{if } x \ge 2 \end{cases}$ .

Which of the following statements is true about the function $f$ at $x=2$ ?

$\lim_{x \to 2} f(x)$ does not exist.
$f(x)$ has a jump discontinuity at $x=2$ .
$f(x)$ has a removable discontinuity at $x=2$ .
$f(x)$ is continuous at $x=2$ . (correct answer)

Explanation: To check for continuity at

x=2

, we must check three conditions: 1.

f(2)

is defined. 2.

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

exists. 3.

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

. \ First,

f(2) = 2(2) + 1 = 5

. \ Second, we check the one-sided limits. The left-hand limit is

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2+1) = 2^2+1 = 5

. The right-hand limit is

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x+1) = 2(2)+1 = 5

. Since the left-hand and right-hand limits are equal,

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

. \ Finally, since

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

, the function is continuous at

x=2

Question 7

A function $f(x)$ is continuous at a point $x=c$ . Which of the following conditions is NOT necessarily required by the definition of continuity at $x=c$ ?

$f(c)$ is defined.
$f(x)$ is differentiable at $x=c$ . (correct answer)
$\lim_{x \to c} f(x)$ exists.
$\lim_{x \to c} f(x) = f(c)$ .

Explanation: The definition of continuity for a function

f(x)

at a point

x=c

requires three conditions: 1)

f(c)

must be defined, 2)

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

must exist, and 3)

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

. Differentiability at a point implies continuity, but continuity does not imply differentiability. For example, the function

f(x) = |x|

is continuous at

x=0

but it is not differentiable at

x=0

. Therefore, differentiability is not a necessary condition for continuity.

Question 8

A function $h(x)$ is continuous at $x=4$ and it is known that $h(4) = -1$ .

What is the value of $\lim_{x \to 4} (h(x) + x^2)$ ?

-1
15 (correct answer)
16
Cannot be determined

Explanation: By the definition of continuity, if

h(x)

is continuous at

x=4

, then

\lim_{x \to 4} h(x) = h(4)

. We are given

h(4) = -1

, so

\lim_{x \to 4} h(x) = -1

. Using the properties of limits, the limit of a sum is the sum of the limits:

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

. We can evaluate these limits:

-1 + 4^2 = -1 + 16 = 15

Question 9

Consider the function $f(x) = \frac{x^2 - 1}{|x - 1|}$ .

Which of the following statements about the function $f$ is true?

$f$ is continuous at $x=1$ .
$\lim_{x \to 1} f(x) = 2$ .
$f$ has a removable discontinuity at $x=1$ .
$\lim_{x \to 1^-} f(x) = -2$ . (correct answer)

Explanation: We must analyze the function by considering the definition of

|x-1|

. \ Case 1:

x > 1

. Then

|x-1| = x-1

. So,

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

. The right-hand limit is

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

. \ Case 2:

x < 1

. Then

|x-1| = -(x-1)

. So,

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

. The left-hand limit is

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

. \ Since the left-hand limit (-2) is not equal to the right-hand limit (2), the overall limit

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

does not exist. This means the function has a jump discontinuity at

x=1

and is not continuous there. Option D correctly states the value of the left-hand limit.

Question 10

The function $f(x)$ is continuous on the interval $[-2, 5]$ . It is known that $f(-2)=3$ and $f(5)=-1$ .

Which theorem guarantees that there is a value $c$ in the interval $(-2, 5)$ such that $f(c)=0$ ?

The Squeeze Theorem
The Extreme Value Theorem
The Mean Value Theorem
The Intermediate Value Theorem (correct answer)

Explanation: The Intermediate Value Theorem (IVT) states that if a function

f

is continuous on a closed interval

[a, b]

, then for any value

N

between

f(a)

and

f(b)

, there exists at least one number

c

(a, b)

such that

f(c) = N

. In this case, the function is continuous on

[-2, 5]

f(-2)=3

, and

f(5)=-1

. Since 0 is a value between -1 and 3, the IVT guarantees that there is a value

c

(-2, 5)

such that

f(c)=0

Question 11

Find the value of $\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)}$ .

0
1
3 (correct answer)
Does not exist

Explanation: This limit is of the indeterminate form

\frac{0}{0}

. We can use the known limit

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

. We can rewrite the expression by multiplying the numerator and denominator by

3x

and

x

respectively:

\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)} = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot 3x \cdot \frac{x}{\sin(x)} \cdot \frac{1}{x} \right) = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot \frac{x}{\sin(x)} \cdot \frac{3x}{x} \right)

. As

x \to 0

3x \to 0

, so

\frac{\sin(3x)}{3x} \to 1

. Also,

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

. The expression simplifies to

1 \cdot 1 \cdot 3 = 3

Question 12

A function $h(x)$ is continuous at $x=4$ and it is known that $h(4) = -1$ .

What is the value of $\lim_{x \to 4} (h(x) + x^2)$ ?

-1
15 (correct answer)
16
Cannot be determined

Explanation: By the definition of continuity, if

h(x)

is continuous at

x=4

, then

\lim_{x \to 4} h(x) = h(4)

. We are given

h(4) = -1

, so

\lim_{x \to 4} h(x) = -1

. Using the properties of limits, the limit of a sum is the sum of the limits:

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

. We can evaluate these limits:

-1 + 4^2 = -1 + 16 = 15

Question 13

Consider the function $f(x) = \frac{x^2 - 1}{|x - 1|}$ .

Which of the following statements about the function $f$ is true?

$f$ is continuous at $x=1$ .
$\lim_{x \to 1} f(x) = 2$ .
$f$ has a removable discontinuity at $x=1$ .
$\lim_{x \to 1^-} f(x) = -2$ . (correct answer)

Explanation: We must analyze the function by considering the definition of

|x-1|

. \ Case 1:

x > 1

. Then

|x-1| = x-1

. So,

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

. The right-hand limit is

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

. \ Case 2:

x < 1

. Then

|x-1| = -(x-1)

. So,

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

. The left-hand limit is

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

. \ Since the left-hand limit (-2) is not equal to the right-hand limit (2), the overall limit

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

does not exist. This means the function has a jump discontinuity at

x=1

and is not continuous there. Option D correctly states the value of the left-hand limit.

Question 14

Find the value of the limit $\lim_{x \to 9} \frac{\sqrt{x} - 3}{x - 9}$ .

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

Explanation: Substituting

x=9

gives the indeterminate form

\frac{0}{0}

. We can resolve this by multiplying the numerator and denominator by the conjugate of the numerator, which is

\sqrt{x} + 3

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

. We can cancel the

(x-9)

terms, leaving

\lim_{x \to 9} \frac{1}{\sqrt{x} + 3}

. Substituting

x=9

now gives

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

Question 15

Consider the piecewise function defined by $f(x) = \begin{cases} x^2 + 1 & \text{if } x < 2 \\ 2x + 1 & \text{if } x \ge 2 \end{cases}$ .

Which of the following statements is true about the function $f$ at $x=2$ ?

$\lim_{x \to 2} f(x)$ does not exist.
$f(x)$ has a jump discontinuity at $x=2$ .
$f(x)$ has a removable discontinuity at $x=2$ .
$f(x)$ is continuous at $x=2$ . (correct answer)

Explanation: To check for continuity at

x=2

, we must check three conditions: 1.

f(2)

is defined. 2.

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

exists. 3.

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

. \ First,

f(2) = 2(2) + 1 = 5

. \ Second, we check the one-sided limits. The left-hand limit is

\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2+1) = 2^2+1 = 5

. The right-hand limit is

\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x+1) = 2(2)+1 = 5

. Since the left-hand and right-hand limits are equal,

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

. \ Finally, since

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

, the function is continuous at

x=2

Question 16

Find the value of the limit $\lim_{x \to 3} \frac{x^2 - x - 6}{x - 3}$ .

-2
0
5 (correct answer)
Does not exist

Explanation: To evaluate this limit, we first try to substitute

x=3

, which results in the indeterminate form

\frac{0}{0}

. This suggests that

(x-3)

is a factor of the numerator. Factoring the numerator gives

x^2 - x - 6 = (x-3)(x+2)

. The expression becomes

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

. For

x \neq 3

, we can cancel the

(x-3)

terms, leaving

\lim_{x \to 3} (x+2)

. Now we can substitute

x=3

to get

3+2 = 5

Question 17

Find the value of $\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)}$ .

0
1
3 (correct answer)
Does not exist

Explanation: This limit is of the indeterminate form

\frac{0}{0}

. We can use the known limit

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

. We can rewrite the expression by multiplying the numerator and denominator by

3x

and

x

respectively:

\lim_{x \to 0} \frac{\sin(3x)}{\sin(x)} = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot 3x \cdot \frac{x}{\sin(x)} \cdot \frac{1}{x} \right) = \lim_{x \to 0} \left( \frac{\sin(3x)}{3x} \cdot \frac{x}{\sin(x)} \cdot \frac{3x}{x} \right)

. As

x \to 0

3x \to 0

, so

\frac{\sin(3x)}{3x} \to 1

. Also,

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

. The expression simplifies to

1 \cdot 1 \cdot 3 = 3

Question 18

Find $\lim_{x \to -\infty} \frac{3x^2 + 5x}{1 - 2x^2}$ .

- $\frac{3}{2}$ (correct answer)
0
$\frac{3}{2}$
$\infty$

Explanation: To find the limit of a rational function as

x \to -\infty

, we can compare the degrees of the numerator and denominator. Both have a degree of 2. Therefore, the limit is the ratio of the leading coefficients. The leading term in the numerator is

3x^2

, and the leading term in the denominator is

-2x^2

. The limit is

\frac{3}{-2} = -\frac{3}{2}

. The direction

x \to -\infty

does not change this result compared to

x \to \infty

when the degrees are equal.

Question 19

Given that $\lim_{x \to a^-} f(x) = L$ and $\lim_{x \to a^+} f(x) = M$ , where $L \neq M$ . Which of the following must be true?

$f(x)$ has a removable discontinuity at $x=a$ .
$f(a)$ is not defined.
$\lim_{x \to a} f(x)$ does not exist. (correct answer)
$f(x)$ has a vertical asymptote at $x=a$ .

Explanation: The condition for the existence of a limit at a point

x=a

is that the left-hand limit must equal the right-hand limit. The problem states that

L \neq M

, meaning

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

. Therefore, the overall limit

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

does not exist. This describes a jump discontinuity. A removable discontinuity requires the limit to exist. It is possible for

f(a)

to be defined (e.g.,

f(a)=L

f(a)=M

), so B is not necessarily true. This situation describes a jump, not a vertical asymptote, so D is incorrect.

Question 20

Find the value of $\lim_{x \to \infty} \frac{6x^3 - 4x + 1}{8x + 2x^3}$ .

0
$\frac{3}{4}$
3 (correct answer)
$\infty$

Explanation: To find the limit of a rational function as

x \to \infty

, we compare the degrees of the numerator and the denominator. The degree of the numerator is 3, and the degree of the denominator is also 3. Since the degrees are equal, the limit is the ratio of the leading coefficients. The leading term in the numerator is

6x^3

and the leading term in the denominator is

2x^3

. The limit is

\frac{6}{2} = 3

. Alternatively, divide every term by the highest power of

x

, which is

x^3

\lim_{x \to \infty} \frac{6 - \frac{4}{x^2} + \frac{1}{x^3}}{\frac{8}{x^2} + 2}

. As

x \to \infty

, the terms with

x

in the denominator approach 0, leaving

\frac{6-0+0}{0+2} = \frac{6}{2} = 3