# Introduction to limits and differentiation

> IB Mathematics: Analysis and Approaches SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-introduction-to-limits-and-differentiation/

This subtopic introduces the core foundation of calculus: limits, the relationship between average and instantaneous rate of change, and differentiation from first principles, the basis for all further differentiation work in IB AA SL.

**Prerequisites:** [Functions notation and polynomial functions](https://www.owlsprep.com/study/ib-math-aa-sl-polynomial-functions/); Gradient of a straight line

## Learning objectives

- Understand the concept of a limit in the context of tangent gradients
- Differentiate simple polynomial functions from first principles
- Interpret the derivative as the instantaneous rate of change and tangent gradient
- Find the equation of a tangent line at a point on a curve

## Limits and the Gradient of a Tangent

**Limit of a function** — 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.

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

*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:
2. $$\frac{f(2+h) - f(2)}{h}$$
3. 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 $h$:
6. 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.

> **tip**
>
> Always simplify the secant gradient expression before taking the limit, to eliminate the $h$ in the denominator.

> **Exam tip:** Marks are awarded for showing how the limit is approached, not just the final answer.

*Calculator:* forbidden

## Differentiation from First Principles

**Derivative from First Principles** — 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.

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

*Example:* For $f(x) = 3x$, $f'(x) = 3$, 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 $f(x) = x^2 + 3x$ from first principles.

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

> **Exam tip:** Always show all expansion and simplification steps, marks are awarded for method not just the final derivative.

*Calculator:* forbidden

## Interpreting the Derivative

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 $y = 2x^2 - x$ at $x=1$.

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

**Check your understanding**

Test your interpretation of the derivative:

1. 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

   *Answer:* The car's instantaneous speed at 5 seconds is 10 m/s

   *Why:* Correct! The derivative of distance with respect to time is instantaneous speed.

*Calculator:* forbidden

## Common pitfalls

- **Wrong:** Trying to substitute $h=0$ before cancelling it from the denominator, leading to undefined $\frac{0}{0}$.
  - Why it fails: The function is undefined at $h=0$, but the limit only depends on values approaching 0, not $h=0$ itself.
  - Correct: Always factor $h$ out of the numerator first, cancel it with the denominator, then substitute $h=0$.
- **Wrong:** Incorrectly expanding $(x+h)^n$, for example writing $(x+h)^2 = x^2 + h^2$.
  - Why it fails: Missing the cross term when expanding binomials, leading to an incorrect derivative.
  - Correct: Always use the binomial expansion rule: $(x+h)^2 = x^2 + 2xh + h^2$ for any $x, h$.
- **Wrong:** Stopping at the simplified $\frac{f(x+h)-f(x)}{h}$ step, forgetting to take the limit explicitly.
  - Why it fails: Confusing the secant gradient expression with the final derivative.
  - Correct: Always write the limit step explicitly and substitute $h \to 0$ to get the final derivative.
- **Wrong:** Using $f'(a)$ as the $y$-coordinate when finding the tangent equation at $x=a$.
  - Why it fails: Mixing up the gradient value and the function value at the point.
  - Correct: Calculate gradient as $f'(a)$ and $y$-coordinate as $f(a)$ separately, keep the two values distinct.

## 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 |

## 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.

- [Basic differentiation rules: power, product, quotient, chain rule](https://www.owlsprep.com/study/ib-math-aa-sl-u5-basic-differentiation-rules-power-product/)
- [Differentiation of trigonometric, exponential and logarithmic functions](https://www.owlsprep.com/study/ib-math-aa-sl-u5-differentiation-of-trigonometric-exponential-and/)
- [Stationary points, monotonicity and concavity](https://www.owlsprep.com/study/ib-math-aa-sl-u5-stationary-points-monotonicity-and-concavity/)

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