DP Maths AI SL 5.1 Limits and Continuity

DP Maths AI SL 5.1 Limits and Continuity - Full notes

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DP Maths AI SL | 5.1 Limits and Continuity

DP MATHEMATICS AI · TOPIC 5: CALCULUS

5.1

Limits and Continuity

Full notes for Applications and Interpretation SL

Contents

1	Rates of change in context
2	Average rate of change
3	Instantaneous rate of change
4	The idea of a limit
5	Limits from graphs and tables
6	Long-run values and limits of sequences
7	Limits at infinity and horizontal asymptotes
8	One-sided limits
9	Continuity
10	Discontinuities in real situations
11	The gradient of a tangent as a limit
12	Estimating derivatives numerically
13	Derivative notation
14	Interpreting derivatives and their units
15	Increasing and decreasing functions
16	Limits with technology
17	Applications: velocity and marginal cost
18	Rates of change in science and economics
19	Mixed review: worked problems
20	Investigation: how fast is the water rising?
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the practice exercise.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
calculate average rates of change as gradients of chords
understand instantaneous rate of change as a limit
estimate limits from tables and graphs, including limits at infinity
identify one-sided limits and continuity
interpret discontinuities in context
estimate derivatives numerically
use the notation dy/dx and f′(x) and interpret units
relate the sign of the derivative to increasing and decreasing functions

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can we measure change at a single instant? What does it mean for a quantity to approach a value?

How this topic is examined

In AI, limits are introduced informally to explain the derivative. Questions ask for average rates of change, estimates of a gradient from a table or graph, interpretation of dy/dx in context with units, long-run values (asymptotes), and continuity of piecewise models.

Give units for every rate of change ('m/s', 'people per year') and explain what the value means in the context.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Rates of change in context

How fast is it changing?

Think about it

A car's odometer reads 12 340 km at 9:00 and 12 460 km at 10:30. What was the car's average speed? Does that tell you its speed at exactly 10:00?

A rate of change measures how quickly one quantity changes as another changes: speed (distance per time), growth rate (people per year), marginal cost (dollars per item). On a graph of y against x, the rate of change is the gradient.

2 Average rate of change

Gradient of a chord.

The average rate of change between x = a and x = b is the gradient of the straight line (chord) joining the two points on the graph.

WORKED EXAMPLE 1 Average rate (basic)

The population of a town was 24 500 in 2015 and 29 300 in 2023. Find the average rate of change.

Solution

(29 300 − 24 500)/8M1

= 600 people per yearA1

WORKED EXAMPLE 2 From a function (standard)

h(t) = 30t − 4.9t2. Find the average velocity between t = 1 and t = 3.

Solution

h(1) = 25.1, h(3) = 45.9A1

(45.9 − 25.1)/2 = 10.4 m/sA1

Practice exercise 1

1. f(x) = x3. Find the average rate of change from x = 1 to x = 4.

2. A tank's volume fell from 850 L to 610 L in 12 minutes. Find the average rate of change.

3 Instantaneous rate of change

Gradient of the tangent.

The instantaneous rate of change at a point is the gradient of the tangent to the curve there. It can be found by taking chords over smaller and smaller intervals: their gradients approach the tangent's gradient.

Figure 1. Chords from (1, 1) get closer to the tangent (dashed) as the second point approaches.

Second point x	3	2	1.4	1.1	1.01	1.001
Chord gradient for y = x2	4	3	2.4	2.1	2.01	2.001

4 The idea of a limit

Approaching a value.

We say the limit of f(x) as x approaches a is L, written limx→a f(x) = L, if f(x) gets as close as we like to L when x is close enough to a (but not equal to a). The chord gradients above have limit 2 as the second point approaches x = 1.

WORKED EXAMPLE 3 Numerical limit (standard)

Estimate limx→0 (2x − 1)/x using a table.

Solution

x = 0.1: 0.7177; x = 0.01: 0.6956; x = 0.001: 0.6934M1 A1

Limit ≈ 0.693 (= ln 2)A1

Practice exercise 2

1. Estimate limx→2 (x2 − 4)/(x − 2) with a table.

2. Estimate limx→0 sin x/x (radians).

5 Limits from graphs and tables

Read the trend.

From a graph, follow the curve towards x = a from both sides and see which y-value it approaches. A hole in the graph (an undefined point) does not prevent a limit from existing.

Practice exercise 3

1. A graph approaches y = 5 from both sides at x = 3 but has an open circle at (3, 5) and a filled point at (3, 2). Find limx→3 f(x) and f(3).

6 Long-run values and limits of sequences

What happens eventually?

Many models settle to a long-run value: a cooling drink approaches room temperature; a drug level approaches a steady state; a geometric series with |r| < 1 approaches a finite sum. The long-run value is the limit as time increases without bound.

WORKED EXAMPLE 4 Long-run temperature (standard)

T(t) = 21 + 70(0.9)t. Find the limit of T as t → ∞ and interpret it.

Solution

(0.9)t → 0M1

T → 21 °C: the room temperatureA1 R1

Practice exercise 4

1. Find the long-run value of P(t) = 500/(1 + 9e−0.3t).

2. Find the limit of un = (3n + 1)/(n + 2) as n → ∞.

7 Limits at infinity and horizontal asymptotes

Behaviour for large x.

If f(x) → L as x → ∞ (or −∞), the line y = L is a horizontal asymptote. For rational functions, compare the leading terms; for exponentials, kax → 0 if 0 < a < 1.

Figure 2. The graph approaches the asymptote y = 3.

Practice exercise 5

1. Find limx→∞ (5 − 2e−0.4x).

2. Find the horizontal asymptote of y = (4x − 3)/(2x + 7).

8 One-sided limits

From the left or the right.

limx→a⁻ f(x) is the limit as x approaches a from below; limx→a⁺ f(x) from above. The (two-sided) limit exists only if the two one-sided limits are equal. Step functions, such as postage costs, have different one-sided limits at the steps.

Practice exercise 6

1. Postage costs $2 up to 100 g and $3.50 from 100 g to 250 g. Find the one-sided limits of the cost as the mass approaches 100 g.

9 Continuity

Drawing without lifting the pen.

A function is continuous at x = a if f(a) exists, the limit exists, and they are equal. Polynomials, exponentials and sinusoidal functions are continuous everywhere. Rational functions are discontinuous where the denominator is zero.

WORKED EXAMPLE 5 Making a piecewise model continuous (standard)

C(x) = 4x + 10 for 0 ≤ x ≤ 5 and 3x + k for x > 5. Find k so that C is continuous.

Solution

At x = 5: 4(5) + 10 = 30M1

3(5) + k = 30: k = 15A1

Practice exercise 7

1. Where is f(x) = 1/(x − 4) discontinuous?

2. f(x) = x2 for x < 2 and ax − 2 for x ≥ 2. Find a for continuity.

10 Discontinuities in real situations

Jumps and gaps.

Real models are often discontinuous: tax brackets with step changes in rates, parking charges per hour, the number of items in stock after a delivery. Interpreting the jump (its size and its location) is part of the modelling.

Figure 3. A step function for parking charges.

11 The gradient of a tangent as a limit

Defining the derivative.

The derivative f′(a) is the limit of chord gradients as the interval shrinks to zero. It is the instantaneous rate of change at x = a, and the gradient of the tangent there.

12 Estimating derivatives numerically

Small intervals.

With a table of values or a function, estimate f′(a) using a small h: the forward difference (f(a + h) − f(a))/h, or more accurately the central difference (f(a + h) − f(a − h))/(2h).

WORKED EXAMPLE 6 Central difference (standard)

Estimate the gradient of y = ex at x = 1 with h = 0.01.

Solution

(e1.01 − e0.99)/0.02M1

= 2.7183 (exact e = 2.71828)A1

t (s)	0	1	2	3	4
d (m)	0	4.2	16.5	37.4	66.1

WORKED EXAMPLE 7 From a table (standard)

Estimate the velocity at t = 2 from the table.

Solution

(37.4 − 4.2)/(3 − 1)M1

= 16.6 m/sA1

Practice exercise 8

1. Estimate the gradient of y = x3 at x = 2 with h = 0.001.

2. Estimate the velocity at t = 3 from the table.

13 Derivative notation

dy/dx and f′(x).

If y = f(x), the derivative is written dy/dx, f′(x), or d/dx(f(x)). With other variables: dV/dt is the rate of change of V with respect to t; dC/dx is the marginal cost. The derivative is itself a function (the gradient function).

14 Interpreting derivatives and their units

Meaning in context.

WORKED EXAMPLE 8 Interpretation (standard)

P(t) is the population of fish (thousands) t years after 2020. Interpret P′(3) = −1.4.

Solution

In 2023, the population is decreasingR1

at a rate of 1400 fish per yearA1

Practice exercise 9

1. V(t) litres is the volume in a tank after t minutes. Interpret V′(5) = 12.

2. C(x) dollars is the cost of x items. Interpret C′(100) = 4.5.

15 Increasing and decreasing functions

Sign of the gradient.

Where f′(x) > 0 the function is increasing; where f′(x) < 0 it is decreasing; where f′(x) = 0 there may be a maximum or minimum (a turning point).

Figure 4. f decreases where f′ < 0 and increases where f′ > 0.

Practice exercise 10

1. From the figure, state the interval where f is decreasing.

2. At a maximum, what is the sign of f′ just before and just after?

16 Limits with technology

Tables, graphs and derivative tools.

GDC tip

Use a table with x values close to a (for example a ± 0.1, 0.01, 0.001) to estimate limits. Most GDCs have a numerical derivative tool (nDeriv or dy/dx on the graph) that gives f′(a) directly.

17 Applications: velocity and marginal cost

Derivatives in context.

WORKED EXAMPLE 9 Velocity (standard)

s(t) = 2t3 − 5t m. Use the GDC to find the velocity at t = 2.

Solution

nDeriv: 19 m/sA1

WORKED EXAMPLE 10 Marginal cost (exam level)

C(x) = 2000 + 15x + 0.02x2. Estimate the cost of producing the 201st item.

Solution

C′(200) ≈ 15 + 0.04(200) = 23 (or C(201) − C(200) = 23.02)M1 A1

About $23A1

Practice exercise 11

1. Revenue R(x) = 50x − 0.1x2. Find the marginal revenue at x = 100 and interpret.

18 Rates of change in science and economics

The same idea, many names.

Rates of change appear under different names in different subjects: velocity and acceleration in physics; reaction rate in chemistry (change in concentration per second); growth rate in biology; marginal cost, marginal revenue and inflation rate in economics; rate of flow in engineering. In each case the rate is the derivative of one quantity with respect to another, and its units are the units of the first quantity divided by the units of the second.

Quantity	Rate	Units
distance s (m)	velocity ds/dt	m/s
velocity v (m/s)	acceleration dv/dt	m/s2
concentration C (mol/L)	reaction rate dC/dt	mol/L per s
cost C ($)	marginal cost dC/dx	$ per item
population P	growth rate dP/dt	individuals per year

WORKED EXAMPLE 11 Reaction rate (standard)

The concentration of a reactant is C(t) = 0.8e−0.05t mol/L. Estimate the reaction rate at t = 10 s using a central difference with h = 0.1.

Solution

(C(10.1) − C(9.9))/0.2M1

= −0.0243 mol/L per s: the concentration is decreasingA1 R1

WORKED EXAMPLE 12 Acceleration from a table (standard)

Velocities (m/s) at t = 0, 2, 4, 6 s are 0, 7.8, 13.9, 18.2. Estimate the acceleration at t = 4 s.

Solution

(18.2 − 7.8)/(6 − 2)M1

= 2.6 m/s2A1

Practice exercise 12

1. Interpret dP/dt = −250 people per year in context.

2. A company's profit P(x) (thousand $) for x thousand items has P′(20) = 1.6. Interpret this.

3. Estimate the acceleration at t = 2 s from the velocity table above.

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 13 Mixed problem (exam level)

f(x) = 100/(1 + 4e−0.5x). Find f(0), the long-run value, and estimate f′(2).

Solution

f(0) = 20A1

Limit 100A1

GDC: f′(2) ≈ 12.0A1

Practice exercise 13

1. Estimate the gradient of y = ln x at x = 4 using h = 0.01 (central difference).

20 Investigation: how fast is the water rising?

A Paper 3 style exploration.

Water is poured into a conical funnel at a constant 50 cm3/s. The depth h (cm) after t seconds is recorded: t = 0, 2, 4, 6, 8, 10; h = 0, 6.2, 7.8, 8.9, 9.8, 10.6. 1. Find the average rate of change of depth over 0 ≤ t ≤ 2 and 8 ≤ t ≤ 10. 2. Explain why the rate decreases. 3. Estimate dh/dt at t = 6 with a central difference. 4. Fit a power model h = atb and estimate dh/dt at t = 6 from the model. 5. Comment on the reliability of the estimates.

Answers. 1. 3.1 cm/s; 0.4 cm/s. 2. The cone widens, so the same volume raises the depth less. 3. (9.8 − 7.8)/4 = 0.5 cm/s. 4. h ≈ 4.9t1/3: dh/dt ≈ (1/3)(4.9)(6)−2/3 ≈ 0.50 cm/s. 5. Central differences are reasonable with smooth data; the model fit and measurement error affect accuracy.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 8 minutes)

f(x) = x2 − 3x.

(a) Find the average rate of change from x = 1 to x = 4. [2]

(b) Estimate f′(2) using the central difference with h = 0.1. [2]

Markscheme

Part	Answer	Marks
(a)	(4 − (−2))/3	M1
	2	A1
(b)	(f(2.1) − f(1.9))/0.2	M1
	1	A1

Examiner's comments. The central difference is exact for quadratics.

Question 2 (Paper 2 style) (about 10 minutes)

A pond's algae coverage is A(t) = 120/(1 + 11e−0.4t) m2, t in days.

(a) Find A(0). [1]

(b) Find the long-run coverage. [1]

Markscheme

Part	Answer	Marks
(a)	10 m2	A1
(b)	120 m2	A1
(c)	A′(5) ≈ 11.5	A1
	on day 5 the area increases by about 11.5 m2 per day	R1 A1

Examiner's comments. Units are required for full marks.

Question 3 (Paper 1 style) (about 6 minutes)

C(x) = 3x + 20 for 0 ≤ x ≤ 10 and 2x + k for x > 10.

Find k for which C is continuous at x = 10. [2]

Markscheme

Part	Answer	Marks
	3(10) + 20 = 2(10) + k	M1
	k = 30	A1

Examiner's comments. Equate the two pieces at the join.

22 Common misconceptions and links

Students often think…	Actually…
average rate = instantaneous rate	only for straight lines
a hole means no limit	the limit can still exist
derivative units ignored	units of y per unit of x
f′ = 0 always means a maximum	could be a minimum or neither

How 5.1 connects to the rest of the course

Topic	Connection
5.3 Derivatives	rules for differentiating
2.2 Functions	gradients
1.3 Geometric series	sums to infinity
5.5 Integration	accumulation

23 Check your understanding

Section A: multiple choice

1. Average rate of change is the gradient of

A. a chord B. a tangent C. the x-axis D. a normal

2. Instantaneous rate is the gradient of

A. a chord B. the curve's asymptote C. the tangent D. a secant

3. limx→∞ 0.8x =

A. 1 B. 0 C. 0.8 D. ∞

4. dV/dt has units

A. units of V B. units of t C. none D. units of V per unit time

5. f′(x) < 0 means f is

A. increasing B. decreasing C. constant D. zero

Section B: structured

6. Estimate limx→0 (ex − 1)/x. [2]

7. Find the average rate of change of y = 2x from x = 0 to x = 3. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. 1

7. 7/3

24 Summary and topic map

Idea	Remember
Average rate	(f(b) − f(a))/(b − a)
Instantaneous rate	limit of chord gradients
Limit	value approached
Asymptote	limit as x → ±∞
Continuity	f(a) = limit
Derivative	dy/dx; gradient function; units
Numerical	central difference

Before the exam, make sure you…

•give units
•interpret in context
•check both sides of a limit
•use small h for estimates

Limits and Continuity on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 5. Topic map for Limits and Continuity.

25 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. 21

2. −20 L per minute

Practice exercise 2

1. 4

2. 1

Practice exercise 3

1. 5; 2

Practice exercise 4

1. 500

2. 3

Practice exercise 5

1. 5

2. y = 2

Practice exercise 6

1. from below $2; from above $3.50; no two-sided limit

Practice exercise 7

1. x = 4

2. a = 3

Practice exercise 8

1. 12.0000

2. (66.1 − 16.5)/2 = 24.8 m/s

Practice exercise 9

1. at 5 minutes, the volume increases by 12 L per minute

2. the 101st item costs about $4.50

Practice exercise 10

1. −1 < x < 1

2. positive before, negative after

Practice exercise 11

1. 30: the 101st item adds about $30 revenue

Practice exercise 12

1. the population is decreasing by 250 people per year at that time

2. at 20 000 items, each extra thousand items adds about $1600 profit

3. (13.9 − 0)/4 = 3.48 m/s2

Practice exercise 13

1. 0.25

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