Introduction to limits and differentiation

IB Mathematics: Analysis and Approaches SL· Unit 5: Calculus, Topic 1· 15 min read

1. Limits and the Gradient of a Tangent★☆☆☆☆⏱ 5 min

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📘 Definition

Limit of a function

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

A limit describes the value that a function $f (x)$ approaches as the input $x$ approaches a given value $a$ , regardless of the value of $f$ at $x = a$ itself.

Example:

For $f (x) = \frac{x ^{2} - 1}{x - 1}$ , $lim_{x \to 1} f (x) = 2$ even though $f (1)$ is undefined.

Limits allow us to move from calculating the gradient of a secant (average rate of change between two points on a curve) to the gradient of a tangent (instantaneous rate of change at one point). As the second point moves closer to the first, the secant gradient approaches the tangent gradient, which we find by taking the limit as the distance between points approaches zero.

📐 Worked Example

Estimate the gradient of the tangent to $f (x) = x^{2}$ at $x = 2$ by calculating the secant gradient for $h = 1, 0.1, 0.01$ , where $h$ is the distance between the two x-values.

1
Write the general expression for secant gradient:
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$\frac{f(2+h) - f(2)}{h}$
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Calculate and simplify the expression:
4
$f(2+h) = (2+h)^2 = 4 + 4h + h^2, \quad f(2) = 4 \\ \frac{(4 + 4h + h^2) - 4}{h} = \frac{h(4 + h)}{h} = 4 + h \quad (h \neq 0)$
5
Substitute the given values of $h$ :
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For $h = 1$ , gradient = $4 + 1 = 5$ ; for $h = 0.1$ , gradient = $4.1$ ; for $h = 0.01$ , gradient = $4.01$ .
7
Draw the conclusion: as $h$ approaches 0, the gradient approaches 4, so the tangent gradient at $x = 2$ is approximately 4.

Exam tip:

Marks are awarded for showing how the limit is approached, not just the final answer.

2. Differentiation from First Principles★★☆☆☆⏱ 6 min

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📘 Definition

Derivative from First Principles

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

The derivative $f^{'} (x)$ (or gradient function) is the limit of the secant gradient as $h$ approaches 0, giving the instantaneous gradient of the tangent at any point $x$ on the curve.

Example:

For $f (x) = 3 x$ , $f^{'} (x) = 3$ , which matches the constant gradient of a straight line.

Differentiation from first principles is a standard method often examined directly in IB AA SL Paper 1, usually worth 4-5 marks. It is critical to remember the formula and show every step to get full marks.

📐 Worked Example

Differentiate $f (x) = x^{2} + 3 x$ from first principles.

1
Write the first principles formula:
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$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
3
Expand $f (x + h)$ :
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$f(x+h) = (x+h)^2 + 3(x+h) = x^2 + 2xh + h^2 + 3x + 3h$
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Subtract $f (x) = x^{2} + 3 x$ :
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$f(x+h) - f(x) = 2xh + h^2 + 3h = h(2x + h + 3)$
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Divide by $h$ (valid for $h \neq = 0$ ):
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$\frac{f(x+h) - f(x)}{h} = 2x + h + 3$
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Take the limit as $h \to 0$ :
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$f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3$

Exam tip:

Always show all expansion and simplification steps, marks are awarded for method not just the final derivative.

3. Interpreting the Derivative★★☆☆☆⏱ 4 min

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The derivative has two key interpretations: graphically, it is the gradient of the tangent to the curve at a point; practically, it is the instantaneous rate of change of the output variable with respect to the input variable. For example, if $s (t)$ is the position of an object at time $t$ , $s^{'} (t)$ is the object's instantaneous velocity at time $t$ .

📐 Worked Example

Find the equation of the tangent to $y = 2 x^{2} - x$ at $x = 1$ .

1
First find the gradient function $y^{'}$ (we use the power rule here, which you will learn, confirmed by first principles: $y^{'} = 4 x - 1$ )
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Calculate the gradient at $x = 1$ :
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$m = y'(1) = 4(1) - 1 = 3$
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Find the $y$ -coordinate of the point at $x = 1$ :
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$y = 2(1)^2 - 1 = 1, \text{ so the point is } (1, 1)$
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Use the point-gradient formula for a line:
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$y - 1 = 3(x - 1) \implies y = 3x - 2$

✓ Quick check

Test your interpretation of the derivative:

If $f (x)$ is the distance a car travels in $x$ seconds, what does $f^{'} (5) = 10$ mean?
- The car traveled 10 meters in 5 seconds
- The car's instantaneous speed at 5 seconds is 10 m/s
- The average speed between 0 and 5 seconds is 10 m/s
Reveal answer
1 —
Correct! The derivative of distance with respect to time is instantaneous speed.

4. Common Pitfalls

Wrong move:

Trying to substitute $h = 0$ before cancelling it from the denominator, leading to undefined $\frac{0}{0}$ .

Why:

The function is undefined at $h = 0$ , but the limit only depends on values approaching 0, not $h = 0$ itself.

Correct move:

Always factor $h$ out of the numerator first, cancel it with the denominator, then substitute $h = 0$ .

Wrong move:

Incorrectly expanding $(x + h)^{n}$ , for example writing $(x + h)^{2} = x^{2} + h^{2}$ .

Why:

Missing the cross term when expanding binomials, leading to an incorrect derivative.

Correct move:

Always use the binomial expansion rule: $(x + h)^{2} = x^{2} + 2 x h + h^{2}$ for any $x, h$ .

Wrong move:

Stopping at the simplified $\frac{f ( x + h ) - f ( x )}{h}$ step, forgetting to take the limit explicitly.

Why:

Confusing the secant gradient expression with the final derivative.

Correct move:

Always write the limit step explicitly and substitute $h \to 0$ to get the final derivative.

Wrong move:

Using $f^{'} (a)$ as the $y$ -coordinate when finding the tangent equation at $x = a$ .

Why:

Mixing up the gradient value and the function value at the point.

Correct move:

Calculate gradient as $f^{'} (a)$ and $y$ -coordinate as $f (a)$ separately, keep the two values distinct.

5. Quick Reference Cheatsheet

Concept	Formula / Notation	Key Meaning
Limit of $f (x)$	$lim_{x \to a} f (x)$	Value $f (x)$ approaches as $x$ nears $a$
First Principles Derivative	$f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$	Derivative from fundamental limit definition
Derivative at $x = a$	$f^{'} (a)$	Gradient of tangent at $(a, f (a))$
Tangent Line at $x = a$	$y - f (a) = f^{'} (a) (x - a)$	Equation of tangent at the given point

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
First principles differentiation 4 mark question
2023 · 1
Find tangent gradient at given point

Going deeper

ib syllabusIB AA SL Calculus SyllabusOfficial syllabus reference for Unit 5

What's Next

This subtopic lays the foundational groundwork for all calculus you will study in IB AA SL. Understanding how differentiation is built from the limit of secant gradients will help you make sense of all the standard differentiation rules we will cover next, and help you avoid common misconceptions when applying these rules to more complex functions like trigonometric and logarithmic functions. Mastery of differentiation from first principles is also regularly tested as a full method question in Paper 1 exams, usually worth 4-5 marks, so it is important to practice the full step-by-step method before moving on.