IB Mathematics: Applications and Interpretation SL — Topic 5 Calculus

Limits and the Derivative

Approach a value numerically and graphically, then use the same limiting idea to understand instantaneous gradient and rate of change.

AI SL · SL 5.1

Learning goal

Estimate limits from tables and graphs, then interpret a derivative both as the gradient of a curve and as an instantaneous rate of change.

Syllabus link

IB Mathematics AI SL: SL 5.1. Informal understanding of limits and the derivative; formal analytic methods for evaluating limits are not required.

Big idea

A limit describes what a function is approaching. A derivative describes what a gradient or quantity is doing at an instant.

Key relationship

$f'(a)=\displaystyle\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ connects secant gradients with the tangent gradient.

Approaching a value: the idea of a limit

A limit is about nearby behaviour. We ask what value the output is approaching as the input gets closer and closer to a chosen number.

Limit notation

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

This reads: “as $x$ approaches $a$, $f(x)$ approaches $L$.” The value of $f(a)$ itself may be different from $L$, or $f(a)$ may not even be defined.

At AI SL, the emphasis is on understanding limits from tables, graphs and the idea of approaching a value. We are not trying to build a formal algebraic theory of limits.

Estimating a limit from a table and from a graph

Worked example 1: estimating a limit numerically

Estimate

\[\lim_{x\to2}\frac{x^2-4}{x-2}.\]

Substituting $x=2$ gives $\frac00$, so instead we look at values of $x$ very close to 2 from both sides.

$x$	1.9	1.99	1.999	2.001	2.01	2.1
$\frac{x^2-4}{x-2}$	3.9	3.99	3.999	4.001	4.01	4.1

The values approach 4 from both sides, so

\[\boxed{\lim_{x\to2}\frac{x^2-4}{x-2}=4}.\]

Calculator table showing values of the expression approaching 4 as x approaches 2 — The numerical table in the handout makes the “approach from both sides” idea visible.

A limit need not equal the function value

Suppose the graph approaches $y=4$ as $x\to2$, but the filled point at $x=2$ is at $y=1$. Then

\[\lim_{x\to2}f(x)=4\qquad\text{but}\qquad f(2)=1.\]

The limit describes what nearby outputs approach; the function value describes the actual output at the point.

Graph with an open circle at y equals 4 and a filled point at y equals 1 when x equals 2 — The open circle shows the approached value; the filled point shows the actual function value.

When a two-sided limit does not exist

A two-sided limit exists only when the function approaches the same value from the left and from the right.

Different one-sided behaviour

If the left-hand values approach 2 while the right-hand values approach 3, there is no single number that the function approaches from both sides. Therefore the two-sided limit does not exist.

\[\lim_{x\to a^-}f(x)=2,\qquad \lim_{x\to a^+}f(x)=3\quad\Longrightarrow\quad \lim_{x\to a}f(x)\text{ does not exist}.\]

Graph showing different left-hand and right-hand limiting values — For a two-sided limit, approaching from both directions must lead to the same value.

From straight-line gradient to the gradient of a curve

The derivative answers a new question: how steep is a curve at one particular point?

For a straight line, the gradient is constant:

\[m=\frac{\Delta y}{\Delta x}.\]

For a curve, the gradient changes from point to point. At one point we measure the gradient of the tangent to the curve.

Derivative at a point

$f'(a)$ is the gradient of the tangent to $y=f(x)$ at $x=a$.

Comparison of a constant straight-line gradient with a tangent gradient on a curve — A straight line has one gradient; a curve has a different tangent gradient at different points.

The derivative as a gradient function

The derivative is not just a single number. If we differentiate a function, we obtain a new function that tells us the gradient everywhere.

Worked example 2: using a derivative function

For $f(x)=x^2$, the derivative is

\[f'(x)=2x.\]

Therefore:

Point	$x=-2$	$x=0$	$x=3$
Gradient $f'(x)$	-4	0	6

The derivative graph $y=f'(x)=2x$ records the tangent gradient of $y=f(x)=x^2$ at every $x$-value.

Graphs of f of x equals x squared and its derivative f prime of x equals 2x — The value of $f'(x)$ tells us the gradient of $f$ at the same $x$-coordinate.

Limits connect secants to a tangent

To estimate a curve's gradient at $x=a$, start with a second nearby point at $x=a+h$. The line joining the two points is a secant.

\[\text{secant gradient}=\frac{f(a+h)-f(a)}{h}.\]

As $h$ becomes smaller, the second point moves towards the first. The secant approaches the tangent, so the secant gradient approaches the derivative:

\[\boxed{f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}}.\]

Sequence of secant lines with smaller h approaching a tangent line — Large $h$ gives a visibly different secant. As $h\to0$, the secant line approaches the tangent.

Worked example 3: estimating the gradient of $f(x)=x^2$ at $x=1$

Use the secant-gradient expression with $a=1$:

\[ \frac{f(1+h)-f(1)}{h} =\frac{(1+h)^2-1}{h} =\frac{2h+h^2}{h} =2+h. \]

For smaller positive values of $h$, the gradients become:

$h$	1	0.5	0.1	0.01	0.001
Secant gradient $2+h$	3	2.5	2.1	2.01	2.001

The values approach 2, so $f'(1)=2$.

Video: What is a derivative?

Use this lesson to revisit the visual journey from secant gradients to the gradient of the tangent and the meaning of the derivative.

Finding a derivative at a point with the TI-Nspire CX

AI Mathematics makes deliberate use of technology. When a numerical derivative is required, the calculator can evaluate $f'(a)$ directly.

Calculator method

Define or enter the function.
Use the numerical derivative command or derivative tool.
Evaluate the derivative at the required $x$-value.
Interpret the numerical result as the gradient of the tangent, with appropriate units when the variables have units.

Video: Derivative at any point on the TI-Nspire CX

The calculator tutorial demonstrates three practical ways of obtaining a derivative value and connects the output back to the tangent gradient.

The derivative as a rate of change

The same derivative idea applies whenever one quantity changes with respect to another. Gradient becomes a rate.

Simple rate model

If a babysitter's total charge after $t$ hours is $C(t)=10t$ euros, then

\[C'(t)=10.\]

The derivative has units euros per hour, so the charge is increasing at a constant rate of €10 per hour.

Table and graph showing a babysitting cost increasing by 10 euros per hour — A constant derivative appears as a straight-line graph with constant gradient.

Worked example 4: average velocity approaching instantaneous velocity

A particle has displacement $s(t)=t^2$ metres. The average velocity from $t=3$ to $t=4$ is

\[\frac{s(4)-s(3)}{4-3}=\frac{16-9}{1}=7\text{ m/s}.\]

Now shorten the time interval:

Interval	[3,4]	[3,3.5]	[3,3.1]	[3,3.01]	[3,3.001]
Average velocity	7	6.5	6.1	6.01	6.001

As the second time approaches 3, the average velocity approaches $6\text{ m/s}$. Therefore the instantaneous velocity at $t=3$ is

\[s'(3)=6\text{ m/s}.\]

Derivative notation and units

The derivative can be written in several equivalent ways. The notation often reflects the variables in the problem.

Function or context	Derivative notation	Meaning
$y=f(x)$	$f'(x)$ or $\frac{dy}{dx}$	change in output per unit change in $x$
Displacement $s(t)$	$\frac{ds}{dt}$	velocity
Volume $V(r)$	$\frac{dV}{dr}$	change in volume per unit change in radius
Cost $C(q)$	$C'(q)$ or $\frac{dC}{dq}$	change in cost per unit change in quantity

Units of a derivative

Derivative units are

\[\frac{\text{units of the vertical quantity}}{\text{units of the horizontal quantity}}.\]

For example, if volume is measured in $\mathrm{cm^3}$ and radius in cm, then $dV/dr$ has units $\mathrm{cm^2}$.

Worked example 5: interpreting a derivative in context

Suppose $V(t)$ is the volume of water in a tank, in litres, after $t$ minutes, and

\[V'(12)=-3.5.\]

The negative sign means the volume is decreasing. The magnitude gives the rate.

Interpretation After 12 minutes, the volume of water is decreasing at $3.5\text{ L/min}$.

Two-part lesson summary

Summary diagram connecting limits, tangent gradients and rates of change — The lesson moves from approaching a value to approaching a tangent gradient, then interprets that derivative as a rate of change.

Part I: limits

A limit describes what outputs approach near an input value.
Approach from both sides for a two-sided limit.
The limit can differ from the actual function value.

Part II: derivatives

$f'(a)$ is the tangent gradient at $x=a$.
$f'(x)$ is the gradient function.
Derivative units are output-units per input-unit.

Practice

These questions follow the progression of the handout: limits first, then derivative meaning, secant gradients, calculator use and contextual interpretation.

A. Limits from tables and graphs

A table of values for $g(x)$ near $x=3$ approaches 5 from both sides. State $\lim_{x\to3}g(x)$.
As $x\to4^-$, $f(x)\to-1$, while as $x\to4^+$, $f(x)\to2$. Does $\lim_{x\to4}f(x)$ exist? Explain.
A graph has an open circle at $(2,6)$ and a filled point at $(2,1)$. State $\lim_{x\to2}f(x)$ and $f(2)$.
Use nearby values to estimate $\displaystyle\lim_{x\to1}\frac{x^2-1}{x-1}$.

B. Derivative and gradient

A graph passes through $(3,7)$, and the tangent at that point has gradient $-2$. State $f(3)$ and $f'(3)$.
Given $f'(x)=3x-4$, find the gradients of $f$ at $x=-2$, $x=0$ and $x=5$.
Explain the difference between $f(4)$ and $f'(4)$.
For $f(x)=x^2$, find the gradient of the secant joining the points with $x=2$ and $x=2.5$.
For $f(x)=x^2$, secant gradients from $x=2$ to $x=2+h$ approach 4 as $h\to0$. Estimate $f'(2)$.
Use a TI-Nspire numerical derivative to estimate $g'(2)$ for $g(x)=x^3-2x+1$. State the geometric meaning of your answer.

C. Rates and units

A car has displacement $s(t)$ km, where $t$ is in hours. Explain the meanings of $\frac{s(5)-s(3)}{5-3}$ and $s'(4)$.
$V'(8)=12$, where $V$ is volume in litres and $t$ is time in minutes. Interpret the statement.
$T'(15)=-0.7$, where $T$ is temperature in °C and time is in minutes. Interpret the statement.
If $V$ is measured in $\mathrm{cm^3}$ and $r$ in cm, give the units of $dV/dr$.
Revenue $R(t)$ is measured in dollars and $t$ in hours. Give the units of $R'(t)$, and explain what $R'(t)>0$ means.

Concise answer key

1. 5

2. No; the one-sided limits are different.

3. Limit = 6; $f(2)=1$.

4. 2

5. $f(3)=7$, $f'(3)=-2$.

6. $-10,-4,11$.

7. $f(4)$ is the output; $f'(4)$ is the tangent gradient.

8. 4.5

9. 4

10. $g'(2)=10$; the tangent gradient is 10.

11. Average velocity from 3 h to 5 h; instantaneous velocity at 4 h, both in km/h.

12. Volume increasing at 12 L/min.

13. Temperature decreasing at 0.7 °C/min.

14. $\mathrm{cm^2}$.

15. $/h; revenue is increasing.

Function or context	Derivative notation	Meaning
\(y=f(x)\)	\(f'(x)\) or \(\frac{dy}{dx}\)	change in output per unit change in \(x\)
Displacement \(s(t)\)	\(\frac{ds}{dt}\)	velocity
Volume \(V(r)\)	\(\frac{dV}{dr}\)	change in volume per unit change in radius
Cost \(C(q)\)	\(C'(q)\) or \(\frac{dC}{dq}\)	change in cost per unit change in quantity

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