IB Maths AA HL · Unit 5: Calculus
IB Maths AA HL First Principles Questions
Exam-style IB Maths AA HL first principles questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 28 questions
- Paper 1: 22
- Paper 2: 6
- 7 easy
- 11 medium
- 9 hard
- 1 starter
- 3 worked examples
Practise First Principles questions →
AA HL formula booklet
What's examined in AA HL first principles
The question bank covers these first principles question types (number of questions in brackets):
- Limit Evaluation Techniques (18)
- Derivative from Definition (6)
- Function Properties & Derivatives (4)
First Principles worked examples
Worked example 1: Applying L'Hôpital's rule · easy
Use L'Hôpital's rule to find the exact value of the limit $\lim_{x \to 0} \frac{e^{3x} - 1}{\sin x}$.
1. Test the limit by direct substitution of $x=0$: $\frac{e^0 - 1}{\sin 0} = \frac{1 - 1}{0} = \frac{0}{0}$.
2. State that because this evaluates to the indeterminate form $\frac{0}{0}$, L'Hôpital's Rule can be applied.
3. Differentiate the numerator with respect to $x$: $\frac{d}{dx}(e^{3x} - 1) = 3e^{3x}$.
4. Differentiate the denominator with respect to $x$: $\frac{d}{dx}(\sin x) = \cos x$.
5. Form the new limit fraction: $\lim_{x \to 0} \frac{3e^{3x}}{\cos x}$.
6. Evaluate the new limit by substituting $x=0$: $\frac{3e^0}{\cos 0} = \frac{3(1)}{1} = \mathbf{3}$.
Examiner tip: You must explicitly state and show that the initial limit evaluates to the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ on your paper to mathematically justify applying L'Hôpital's Rule for the method mark.
Worked example 2: Derivative from first principles · medium
Use the formal definition of the derivative from first principles to prove that if $f(x) = x^3$, then $f'(x) = 3x^2$.
1. State the formal limit definition of the derivative: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$.
2. Substitute the specific function into the formula: $f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}$.
3. Expand the binomial cubic correctly: $f'(x) = \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$.
4. Cancel the $x^3$ terms in the numerator: $f'(x) = \lim_{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$.
5. Factor out and cancel the common $h$ term: $f'(x) = \lim_{h \to 0} (3x^2 + 3xh + h^2)$.
6. Evaluate the limit by directly substituting $h=0$: $3x^2 + 3x(0) + (0)^2 = \mathbf{3x^2}$.
Examiner tip: Never drop the "$\lim_{h \to 0}$" notation in your working steps until the very final line when you actually evaluate the limit by setting $h=0$. Dropping it early is a notation error penalized by examiners.
Worked example 3: Evaluating limits with Maclaurin series · hard
Evaluate exactly $\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$ using L'Hôpital's Rule or Maclaurin series substitution.
1. Recognize that applying L'Hôpital's rule would require differentiating four times, so substituting the Maclaurin series for $\cos x$ is much faster.
2. State the standard Maclaurin expansion for $\cos x$: $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$
3. Substitute this expansion into the numerator of the limit: $\lim_{x \to 0} \frac{\left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots\right) - 1 + \frac{x^2}{2}}{x^4}$.
4. Cancel the constant $1$ and the $x^2$ terms perfectly: $\lim_{x \to 0} \frac{\frac{x^4}{24} - \frac{x^6}{720} + \dots}{x^4}$.
5. Divide every term in the numerator by the denominator $x^4$: $\lim_{x \to 0} \left( \frac{1}{24} - \frac{x^2}{720} + \dots \right)$.
6. Evaluate the limit by letting $x$ approach zero, annihilating all higher-order terms: $\mathbf{\frac{1}{24}}$.
Examiner tip: Using a Maclaurin series expansion to evaluate a limit is almost always faster and less error-prone than applying L'Hôpital's rule sequentially for high-power polynomial denominators, which gets very messy.
Try these IB Maths AA HL first principles questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 3 marks · Paper 1
Evaluate the following limit, showing your working clearly:
\[\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\]
Question 2 · medium · 5 marks · Paper 2
A function \(f(x)\) is defined as:
\[f(x) = \frac{\ln x}{x^2 - 1}\]
The function is undefined at \(x = 1\). Determine the value that \(f(1)\) must be assigned in order to make the function continuous at \(x = 1\).
Question 3 · hard · 6 marks · Paper 1
Evaluate the following limit requiring repeated applications of L’Hôpital’s rule:
\[\lim_{x \to 0} \frac{x - \tan x}{x^3}\]
All 28 first principles questions with mark schemes →
FAQ
How many IB Maths AA HL first principles questions are there?
There are 28 exam-style first principles questions in the AA HL question bank (Paper 1: 22 · Paper 2: 6), graded 7 easy, 11 medium, 9 hard, 1 starter. Every question has a full IB-style mark scheme (M, A and R marks).
Is first principles on Paper 1 or Paper 2?
Both. In the bank, Paper 1: 22 · Paper 2: 6. Paper 1 is non-calculator, so practise the exact-value algebra as well as the GDC methods.
Where can I get the mark schemes?
Open the AA HL Unit 5 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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