IB Mathematics: Applications and Interpretation Quiz: Limits And Continuity

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Limits And ContinuityQuestion 1 of 20

A parking garage charges for parking based on the number of hours, $h$ . The cost, $C(h)$ , is $5 for the first hour or any fraction thereof, and an additional $3 for each subsequent hour or fraction thereof. This can be modelled by a step function where the cost for $h $hours is$ C(h) = 5 $for$ 0 < h \le 1 $,$ C(h) = 8 $for$ 1 < h \le 2$, and so on.

Calculate the value of $\lim_{h \to 2^+} C(h) - \lim_{h \to 2^-} C(h)$ .

-3

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Limits And Continuity

Practice Limits And Continuity in IB Mathematics: Applications and Interpretation 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: Applications and Interpretation.

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Question 1

Calculate the value of $\lim_{h \to 2^+} C(h) - \lim_{h \to 2^-} C(h)$ .

-3
0
3 (correct answer)
8

Explanation: First, we find the limit from the right,

\lim_{h \to 2^+} C(h)

. For values of

h

slightly greater than 2 (e.g., 2.1 hours), the car is in its third hour of parking. The cost for

2 < h \le 3

would be

8+3=11

. So,

\lim_{h \to 2^+} C(h) = 11

. Second, we find the limit from the left,

\lim_{h \to 2^-} C(h)

. For values of

h

slightly less than 2 (e.g., 1.9 hours), the car is in its second hour of parking, so the cost is

8

. Thus,

\lim_{h \to 2^-} C(h) = 8

. The difference is

11 - 8 = 3

Question 2

The concentration of a drug in a patient's bloodstream, $C$ (in mg/L), $t$ hours after injection is modelled by the function $C(t) = \frac{120t}{t^2 + 4}$ .

What does the model predict for the long-term concentration of the drug in the bloodstream as time increases indefinitely?

0 mg/L (correct answer)
30 mg/L
60 mg/L
120 mg/L

Explanation: To find the long-term concentration, we evaluate the limit as

t \to \infty

. For the rational function

C(t) = \frac{120t}{t^2 + 4}

, the degree of the denominator (2) is greater than the degree of the numerator (1). Therefore, the limit as

t

approaches infinity is 0. This means the drug concentration approaches zero as the drug is eventually eliminated from the patient's system.

Question 3

The population of an insect colony, in thousands, is modelled by the function $P(t) = 0.5t^2 + 8$ , where $t$ is the number of months. At exactly $t=6$ months, a predator is introduced, causing an immediate drop in the insect population. The population is measured to be 24 thousand right after the introduction.

Which statement correctly describes the population model at $t=6$ ?

The model is continuous at $t=6$ with $P(6)=24$ .
The model has $\lim_{t \to 6} P(t) = 26$ .
The model has $\lim_{t \to 6^-} P(t) = 26$ and $P(6)=24$ . (correct answer)
The model has $\lim_{t \to 6^-} P(t) = 24$ and $P(6)=26$ .

Explanation: The limit from the left,

\lim_{t \to 6^-} P(t)

, represents the population just before the predator is introduced. Using the original model, this is

0.5(6^2) + 8 = 0.5(36) + 8 = 18 + 8 = 26

thousand. The passage states that the measured population at

t=6

(i.e., just after the event) is 24 thousand, so

P(6)=24

. Because the limit from the left (26) does not equal the function's value (24), the function is discontinuous. Statement C correctly identifies both values.

Question 4

The distance travelled by a cyclist is given by the function $s(t) = t^2 + 2t$ , where $s$ is in metres and $t$ is in seconds.

What is the physical interpretation of the mathematical expression $\lim_{h \to 0} \frac{s(5+h)-s(5)}{h}$ ?

The total distance travelled by the cyclist in the first 5 seconds.
The average speed of the cyclist during the first 5 seconds.
The acceleration of the cyclist at exactly 5 seconds.
The instantaneous speed of the cyclist at exactly 5 seconds. (correct answer)

Explanation: The expression

\frac{s(5+h)-s(5)}{h}

represents the average rate of change of distance (average speed) over a small time interval

h

starting at

t=5

. The limit as this interval

h

approaches zero gives the instantaneous rate of change of distance at

t=5

. The instantaneous rate of change of distance is the definition of instantaneous speed (or velocity).

Question 5

The average cost, $AC$ , in dollars per unit to produce $x$ units of a product is given by the function $AC(x) = \frac{12x + 4800}{x}$ , where $x > 0$ .

Which statement correctly interprets the value of $\lim_{x \to 0^+} AC(x)$ ?

As production begins, the average cost per unit approaches $12.
As production begins, the average cost per unit approaches $4800.
As production begins, the average cost per unit becomes infinitely large. (correct answer)
As production begins, the average cost per unit approaches $0.

Explanation: We are evaluating the limit of

AC(x)

x

approaches 0 from the positive side. As

x \to 0^+

, the numerator

12x + 4800

approaches 4800, and the denominator

x

approaches 0. A constant divided by a very small positive number results in a very large positive number. Thus,

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

. This means the fixed costs are spread over very few units, making the average cost per unit extremely high.

Question 6

A function, $B(t)$ , represents the balance in a person's bank account over a period of 30 days, where $t$ is the number of days. The account balance changes due to various transactions.

Which of the following financial events would be best modelled by a jump discontinuity in the function $B(t)$ ?

Interest being compounded continuously on the account.
The bank processing a deposited cheque at a specific time. (correct answer)
A forecast of future spending for the next week.
The gradual decrease in purchasing power due to inflation.

Explanation: A jump discontinuity represents an instantaneous change in value. A cheque being deposited and processed causes the account balance to jump up instantly at a specific point in time. Continuous interest (A) results in a smooth, continuous increase. A forecast (C) is a prediction, not a change in the actual balance function. Inflation (D) is a gradual, continuous process.

Question 7

The effort $E$ required to memorize a list of $n$ items is modelled by the function $E(n) = \frac{2n^2}{n-5}$ for $n>5$ .

Which statement correctly interprets the meaning of $\lim_{n \to 5^+} E(n)$ ?

The effort to memorize 5 items is 0.
The effort to memorize slightly more than 5 items becomes infinitely large. (correct answer)
The effort to memorize slightly more than 5 items approaches 10.
The effort to memorize slightly more than 5 items becomes a large negative number.

Explanation: We are evaluating the limit as

n

approaches 5 from the right (positive) side. As

n \to 5^+

, the numerator

2n^2

approaches

2(5^2) = 50

. The denominator

n-5

approaches 0 from the positive side (e.g., 5.01 - 5 = 0.01). A positive constant divided by a very small positive number results in a very large positive number. Therefore, the limit is

+\infty

. This suggests the model is not valid at or very near

n=5

, as it predicts an infinite effort.

Question 8

The proficiency, $P(t)$ , of a new employee on a task is measured as a percentage and is modelled by $P(t) = 100 - 80(0.85)^t$ , where $t$ is the number of weeks of training.

What is the interpretation of $\lim_{t \to \infty} P(t)$ in the context of this model?

The employee's initial proficiency before any training is 100%.
The employee will eventually achieve a maximum proficiency of 100%. (correct answer)
The employee's proficiency will eventually decrease to 80%.
The employee's proficiency never exceeds 85%.

Explanation: We need to evaluate the limit of

P(t)

t \to \infty

. Since the base of the exponential term is 0.85 (which is between 0 and 1), the term

(0.85)^t

approaches 0 as

t

becomes very large. Therefore,

\lim_{t \to \infty} P(t) = 100 - 80(0) = 100

. This means that with enough training, the model predicts the employee's proficiency will approach a maximum of 100%.

Question 9

At precisely midnight on January 1st, the price of a software subscription is scheduled to increase from $10 per month to $12 per month. Let $P(t) $be the function representing the price of the subscription, where$ t$ is time.

How is the discontinuity in the price function $P(t)$ at midnight best classified?

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

Explanation: When analyzing functions that represent real-world scenarios with sudden changes, you need to identify what type of discontinuity occurs at the point of change. Discontinuities are classified based on the behavior of the function's left and right limits. At midnight, the subscription price jumps instantly from $10 to $12. This creates a situation where the left-hand limit (approaching midnight from before) is $

\lim_{t \to 0^-} P(t) = 10

, while the right-hand limit (approaching midnight from after) is

\lim_{t \to 0^+} P(t) = 12

$. Since these one-sided limits exist but are not equal, this is a jump discontinuity, making D correct. Let's examine why the other options don't apply. Option A (removable discontinuity) occurs when the left and right limits exist and are equal, but the function value at that point is either undefined or different from the limit—you could "remove" the discontinuity by redefining the function at that single point. That's not the case here since the limits aren't equal. Option B (infinite discontinuity) happens when at least one of the one-sided limits approaches infinity, which clearly doesn't occur with prices jumping from $10 to $12. Option C (continuous) would require the left limit, right limit, and function value to all be equal at midnight, but we have a clear price jump. For IB exam success, remember that jump discontinuities are common in piecewise functions representing real-world situations like pricing changes, tax brackets, or shipping rates—look for sudden "jumps" between finite values.

Question 10

A scientist is measuring the response, $R(x)$ , of a chemical reaction to a catalyst added in quantity $x$ . The following table shows measurements for $x$ values close to 3 ml. | $x$ (ml) | 2.9 | 2.99 | 2.999 | 3.001 | 3.01 | 3.1 | |---|---|---|---|---|---|---| | $R(x)$ | 14.71 | 14.970 | 14.997 | 15.003 | 15.030 | 15.31 |

Based on the numerical data, what is the best estimate for $\lim_{x \to 3} R(x)$ ?

15.0 (correct answer)
14.997
14.71
The limit does not exist.

Explanation: When you encounter a table of values approaching a specific point, you're being asked to estimate a limit using numerical evidence. This tests your understanding of limits as the behavior of a function as the input approaches a particular value. To find

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

, examine what happens to the function values as

x

gets closer and closer to 3 from both sides. Looking at the table, as

x

approaches 3 from the left (2.9 → 2.99 → 2.999),

R(x)

approaches 15: the values are 14.71, 14.970, and 14.997. From the right (3.001 ← 3.01 ← 3.1),

R(x)

also approaches 15: the values are 15.003, 15.030, and 15.31. The values closest to

x = 3

(at 2.999 and 3.001) give

R(x)

values of 14.997 and 15.003, both extremely close to 15. Answer A (15.0) is correct because both one-sided limits approach this value, indicating the limit exists and equals 15. Answer B (14.997) is wrong because this is just one specific function value, not the limit that both sides approach. Answer C (14.71) is incorrect because this value occurs at

x = 2.9

, which is relatively far from 3 and doesn't represent the limiting behavior. Answer D is wrong because the limit clearly exists—both sides approach the same value. Study tip: For numerical limit problems, always check values from both sides of the target point. The limit exists when both sides approach the same number, regardless of what happens exactly at that point.

Question 11

The concentration of a drug in a patient's bloodstream, $C$ (in mg/L), $t$ hours after injection is modelled by the function $C(t) = \frac{120t}{t^2 + 4}$ .

What does the model predict for the long-term concentration of the drug in the bloodstream as time increases indefinitely?

0 mg/L (correct answer)
30 mg/L
60 mg/L
120 mg/L

Explanation: To find the long-term concentration, we evaluate the limit as

t \to \infty

. For the rational function

C(t) = \frac{120t}{t^2 + 4}

, the degree of the denominator (2) is greater than the degree of the numerator (1). Therefore, the limit as

t

approaches infinity is 0. This means the drug concentration approaches zero as the drug is eventually eliminated from the patient's system.

Question 12

The average cost, $AC$ , in dollars per unit to produce $x$ units of a product is given by the function $AC(x) = \frac{12x + 4800}{x}$ , where $x > 0$ .

Which statement correctly interprets the value of $\lim_{x \to 0^+} AC(x)$ ?

As production begins, the average cost per unit approaches $12.
As production begins, the average cost per unit approaches $4800.
As production begins, the average cost per unit becomes infinitely large. (correct answer)
As production begins, the average cost per unit approaches $0.

Explanation: We are evaluating the limit of

AC(x)

x

approaches 0 from the positive side. As

x \to 0^+

, the numerator

12x + 4800

approaches 4800, and the denominator

x

approaches 0. A constant divided by a very small positive number results in a very large positive number. Thus,

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

. This means the fixed costs are spread over very few units, making the average cost per unit extremely high.

Question 13

A computer simulation models a particle's velocity, $v$ in m/s, using the function $v(t) = \frac{2t^2 - 18}{t-3}$ for time $t \ne 3$ seconds. Due to a computational limitation, the function is undefined at $t=3$ .

To create a continuous velocity model, what value should be assigned to $v(3)$ ?

0 m/s
6 m/s
12 m/s (correct answer)
18 m/s

Explanation: For continuity at

t=3

, we need

v(3)

to equal the limit as

t

approaches 3. We can simplify the expression by factoring:

v(t) = \frac{2(t^2 - 9)}{t-3} = \frac{2(t-3)(t+3)}{t-3} = 2(t+3)

for

t \ne 3

. Therefore,

\lim_{t \to 3} v(t) = 2(3+3) = 12

. Setting

v(3) = 12

ensures the velocity model is continuous.

Question 14

A city's water tariff, $C(v)$ , is the cost in dollars for consuming $v$ cubic metres ( $m^3$ ) of water in a month. The cost is modelled by the piecewise function: $C(v) = \begin{cases} 2v & \text{if } 0 \le v < 20 \\ k + 1.5v & \text{if } v \ge 20 \end{cases}$ . The city council wants to set the parameter $k$ so that the cost function is continuous for all consumption levels.

What must be the value of the constant $k$ for the cost function $C(v)$ to be continuous at $v=20$ ?

10 (correct answer)
20
30
40

Explanation: For the function to be continuous at

v=20

, the limit from the left must equal the limit from the right. We set the two expressions equal to each other at

v=20

\lim_{v \to 20^-} C(v) = \lim_{v \to 20^+} C(v)

. This gives

2(20) = k + 1.5(20)

. So,

40 = k + 30

. Solving for

k

gives

k = 40 - 30 = 10

Question 15

The altitude of a weather balloon is being controlled by a remote operator. Near a critical time $t=10$ seconds, its altitude in metres is modelled by $A(t) = 500 + 4\sin(\frac{1}{t-10})$ .

Which statement best describes the balloon's altitude as time $t$ gets extremely close to 10 seconds?

The altitude approaches a stable height of 500 metres.
The altitude approaches a stable height of 504 metres.
The altitude increases towards infinity.
The altitude oscillates between 496 and 504 metres with increasing frequency. (correct answer)

Explanation: When analyzing functions with extreme behavior near critical points, you need to examine what happens to each component as the variable approaches that value. Here, as

t

approaches 10, the expression

\frac{1}{t-10}

becomes increasingly large in magnitude (approaching

\pm\infty

). The key insight is understanding how

\sin(\frac{1}{t-10})

behaves as

t \to 10

. Since the argument

\frac{1}{t-10}

grows without bound, the sine function oscillates between -1 and +1 with increasing frequency. This means

4\sin(\frac{1}{t-10})

oscillates between -4 and +4, making the complete function

A(t) = 500 + 4\sin(\frac{1}{t-10})

oscillate between 496 and 504 metres. Option A is wrong because the altitude doesn't stabilize at 500 metres—it continuously oscillates around this value. Option B incorrectly suggests the altitude approaches 504 metres, which would only occur if the sine function approached +1, but it oscillates between all values. Option C is incorrect because while the frequency increases toward infinity, the altitude itself remains bounded between 496 and 504 metres due to the sine function's limited range. Option D correctly captures both essential behaviors: the oscillation between the bounded values (496 to 504 metres) and the increasing frequency as

t

approaches 10. Study tip: When you see

\frac{1}{x-a}

inside a trigonometric function near

x = a

, immediately think "oscillating with increasing frequency" rather than approaching a single limit value.

Question 16

The proficiency, $P(t)$ , of a new employee on a task is measured as a percentage and is modelled by $P(t) = 100 - 80(0.85)^t$ , where $t$ is the number of weeks of training.

What is the interpretation of $\lim_{t \to \infty} P(t)$ in the context of this model?

The employee's initial proficiency before any training is 100%.
The employee will eventually achieve a maximum proficiency of 100%. (correct answer)
The employee's proficiency will eventually decrease to 80%.
The employee's proficiency never exceeds 85%.

Explanation: We need to evaluate the limit of

P(t)

t \to \infty

. Since the base of the exponential term is 0.85 (which is between 0 and 1), the term

(0.85)^t

approaches 0 as

t

becomes very large. Therefore,

\lim_{t \to \infty} P(t) = 100 - 80(0) = 100

. This means that with enough training, the model predicts the employee's proficiency will approach a maximum of 100%.

Question 17

How is the discontinuity in the price function $P(t)$ at midnight best classified?

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

\lim_{t \to 0^-} P(t) = 10

, while the right-hand limit (approaching midnight from after) is

\lim_{t \to 0^+} P(t) = 12

Question 18

The monthly cost, $C(g)$ , for a mobile data plan is modelled by the function $C(g) = \begin{cases} 25 & \text{if } 0 < g \le 2 \\ 25 + 15(g-2) & \text{if } 2 < g \le 10 \\ 150 & \text{if } g > 10 \end{cases}$ , where $g$ is the data used in gigabytes.

At which data usage level(s) is the cost function $C(g)$ discontinuous?

At $g=2$ only
At $g=10$ only (correct answer)
At $g=2$ and $g=10$
The function is continuous for all $g > 0$

Explanation: We check the boundary points. At

g=2

C(2)=25

\lim_{g \to 2^-} C(g) = 25

\lim_{g \to 2^+} C(g) = 25 + 15(2-2) = 25

. Since the value and both one-sided limits are 25, the function is continuous at

g=2

. At

g=10

C(10) = 25 + 15(10-2) = 25 + 15(8) = 25+120=145

. The limit from the left is

\lim_{g \to 10^-} C(g) = 145

. The limit from the right is

\lim_{g \to 10^+} C(g) = 150

. Since the left-hand limit (145) does not equal the right-hand limit (150), the function is discontinuous at

g=10

Question 19

The population of a species of fish in a newly formed lake is modelled by the function $P(t) = \frac{4000t + 500}{t+2}$ , where $t$ is the number of years since the lake was formed.

According to the model, what is the carrying capacity of the lake for this species of fish? This is the value the population approaches in the long run.

250
500
4000 (correct answer)
4500

Explanation: The carrying capacity is the limit of the population function as time

t

approaches infinity. For the rational function

P(t) = \frac{4000t + 500}{t+2}

, the degree of the numerator and the denominator are both 1. The limit is the ratio of the leading coefficients, which is

\frac{4000}{1} = 4000

Question 20

A cup of coffee, initially at 90°C, is left to cool in a room with a constant temperature of 22°C. According to Newton's Law of Cooling, its temperature $T$ (in °C) after $t$ minutes is modelled by $T(t) = 22 + 68e^{-0.04t}$ .

What temperature does the coffee approach as it is left to cool for a very long time?

0°C
22°C (correct answer)
68°C
90°C

Explanation: We need to find the limit of the temperature function as

t \to \infty

. As

t \to \infty

, the term

e^{-0.04t}

approaches 0. Therefore,

\lim_{t \to \infty} T(t) = \lim_{t \to \infty} (22 + 68e^{-0.04t}) = 22 + 68(0) = 22

. The coffee's temperature approaches the room temperature.