IB Math AA HL Maclaurin Series & L'Hôpital — Complete Cheatsheet

Maclaurin Series, L'Hôpital's Rule and Limits (AHL 5.19–5.20) form the most concept-heavy section of IB Mathematics Analysis & Approaches HL. The mechanical work — expanding $e^x$, $\sin x$, $\cos x$ and $\ln(1+x)$, applying L'Hôpital, integrating term by term — is straightforward; the difficulty is choosing the right tool fast and meeting the strict notational requirements that IB mark schemes enforce. Forgetting to write "form $0/0$" before applying L'Hôpital, or omitting $\lim_{x \to a}$ on a single line, costs the final A1 every time.

This cheatsheet condenses every formula, validity range, indeterminate form, and exam trap from AHL 5.19–5.20 into one revisable page. It also includes the comparison table that tells you when to use Maclaurin instead of L'Hôpital — a decision that saves 4–5 minutes per Paper 1 question. Scroll to the bottom for the printable PDF and the full notes booklet.

§1 — The Six Standard Maclaurin Series AHL 5.19

The Maclaurin formula

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \cdots$$

Extracting derivatives: if $f(x) = \sum c_n x^n$, then $f^{(n)}(0) = n! \cdot c_n$.

Validity ranges

The six series — must be memorised

§2 — Techniques for Building New Series AHL 5.19

Substitution

Replace $x$ with $g(x)$ in any standard series; adjust the validity condition accordingly.

Validity: if replacing $x \to g(x)$, the original condition becomes $|g(x)| < 1$ (or $\leq 1$).

Multiplication of two series

Collect by power: $(A_0 + A_1 x + A_2 x^2 + \cdots)(B_0 + B_1 x + B_2 x^2 + \cdots)$ gives

• Coefficient of $x^0$: $A_0 B_0$
• Coefficient of $x^1$: $A_0 B_1 + A_1 B_0$
• Coefficient of $x^2$: $A_0 B_2 + A_1 B_1 + A_2 B_0$
• Coefficient of $x^n$: $\sum_{k=0}^{n} A_k B_{n-k}$

Example: $e^x \sin x = x + x^2 + \dfrac{x^3}{3} - \dfrac{x^4}{30} + \cdots$.

Term-by-term differentiation and integration

Differentiate: $\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots \Rightarrow \cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$  ✓

Integrate: $\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - \cdots \Rightarrow \arctan x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$  ✓

Constant of integration: determined by evaluating at $x = 0$.

Composition (nested substitution)

To find a series for $f(g(x))$:

1. Write the series for $g(x)$ up to the needed order.
2. Substitute that series into the series for $f(u)$.
3. Expand powers of $g(x)$ and collect.

Example: $e^{\sin x}$ with $u = \sin x = x - \dfrac{x^3}{6} + \cdots$: $e^u = 1 + u + \dfrac{u^2}{2} + \dfrac{u^3}{6} + \cdots = 1 + x + \dfrac{x^2}{2} + 0\cdot x^3 + \cdots$.

Derived series worth knowing

• $\sec x = 1 + \dfrac{x^2}{2} + \dfrac{5x^4}{24} + \cdots$
• $e^{x^2} = 1 + x^2 + \dfrac{x^4}{2} + \dfrac{x^6}{6} + \cdots$
• $\sinh x = x + \dfrac{x^3}{6} + \dfrac{x^5}{120} + \cdots$
• $\dfrac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots$
• $\dfrac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots$
• $\ln\!\left(\dfrac{1+x}{1-x}\right) = 2x + \dfrac{2x^3}{3} + \dfrac{2x^5}{5} + \cdots$

§3 — L'Hôpital's Rule AHL 5.20

The rule

If $\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)}$ produces the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$, then

$$\lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}$$

provided the right-hand limit exists. Differentiate the numerator and the denominator separately — this is NOT the quotient rule. The $\lim_{x \to a}$ notation must appear at every step.

All indeterminate forms

Step-by-step protocol

1. Substitute $x = a$ — identify the form.
2. State the form: "form $\dfrac{0}{0}$"  (R1 mark in IB).
3. Differentiate the numerator; differentiate the denominator.
4. Substitute $x = a$ again.
5. If still indeterminate, repeat from step 2.
6. Write $\lim_{x \to a}$ at every line until you can substitute.

Key examples of each form

• $0/0$: $\lim_{x \to 0}\dfrac{\sin x}{x} \to \lim_{x \to 0}\dfrac{\cos x}{1} = 1$ ✓
• Two applications: $\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} \to \lim\dfrac{\sin x}{2x} \to \lim\dfrac{\cos x}{2} = \dfrac{1}{2}$ ✓
• $\infty/\infty$: $\lim_{x \to \infty}\dfrac{\ln x}{\sqrt{x}} \to \lim\dfrac{1/x}{1/(2\sqrt{x})} = \lim\dfrac{2}{\sqrt{x}} = 0$ ✓
• $0 \cdot \infty$: $\lim_{x \to 0^+} x \ln x = \lim\dfrac{\ln x}{1/x} \to \lim\dfrac{1/x}{-1/x^2} = \lim(-x) = 0$ ✓

§4 — Using Maclaurin Series to Evaluate Limits AHL 5.19–5.20

The series method for limits at $x = 0$

1. Expand the numerator using a series; keep enough terms to survive cancellation.
2. Expand the denominator if needed.
3. Cancel the common $x^n$ factor from top and bottom.
4. Take $x \to 0$ — the answer is the leading coefficient.

Key insight: $\displaystyle\lim_{x \to 0}\dfrac{f(x)}{x^n}$ equals the coefficient of $x^n$ in the series for $f(x)$, provided all lower-degree terms of $f$ are zero.

Standard limits (know without working)

• $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$     $\displaystyle\lim_{x \to 0}\dfrac{\tan x}{x} = 1$
• $\displaystyle\lim_{x \to 0}\dfrac{1 - \cos x}{x^2} = \dfrac{1}{2}$     $\displaystyle\lim_{x \to 0}\dfrac{e^x - 1}{x} = 1$
• $\displaystyle\lim_{x \to 0}\dfrac{\ln(1 + x)}{x} = 1$     $\displaystyle\lim_{x \to 0}\dfrac{\arctan x}{x} = 1$
• $\displaystyle\lim_{x \to \infty} x^n e^{-x} = 0$     $\displaystyle\lim_{x \to \infty}\dfrac{\ln x}{x^p} = 0$

Maclaurin vs L'Hôpital — when to use each

§5 — Approximate Integration Using Series AHL 5.19

Integrating a series term by term

1. Find the Maclaurin series for the integrand.
2. Keep as many terms as the required accuracy demands.
3. Integrate each term as a polynomial: $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}$.
4. Evaluate at the limits (or leave as a series if indefinite).

Why? Functions like $e^{x^2}$, $\dfrac{\sin x}{x}$, $e^{x^2}\cos(x^2)$ have no elementary antiderivative — the series is the only Paper 1 method.

Useful integrand series

• $\dfrac{\sin x}{x} = 1 - \dfrac{x^2}{6} + \dfrac{x^4}{120} - \cdots$
• $e^{x^2} = 1 + x^2 + \dfrac{x^4}{2} + \dfrac{x^6}{6} + \cdots$
• $\dfrac{\ln(1 + x)}{x} = 1 - \dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + \cdots$
• $\dfrac{e^x - 1}{x} = 1 + \dfrac{x}{2} + \dfrac{x^2}{6} + \cdots$

Accuracy of the approximation

For the approximation $\int_0^a f(x)\,dx \approx \int_0^a P_n(x)\,dx$ where $P_n$ is the degree-$n$ Maclaurin polynomial:

• Alternating series: error $\leq$ first omitted term (evaluated at the upper limit).
• General: more terms $=$ better accuracy; for small $a$ convergence is rapid.
• Warning: for large upper limits (e.g. $\int_0^{\pi/2}$), the Maclaurin truncation may converge slowly — many terms needed for acceptable accuracy.

§6 — Deriving and Applying the Binomial Series AHL 5.19

Generalised binomial series

$$\displaystyle (1 + x)^p = 1 + p x + \frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + \cdots$$

Valid for $|x| < 1$ when $p \notin \mathbb{Z}^+$.

Using differentiation and integration on series

Differentiate $\dfrac{1}{(1-x)^2}$ from $\dfrac{1}{1-x}$: $\dfrac{d}{dx}(1 + x + x^2 + \cdots) = 1 + 2x + 3x^2 + \cdots = \dfrac{1}{(1-x)^2}$ ✓

Integrate $\dfrac{1}{1+x^2}$ to get $\arctan x$: $\int (1 - x^2 + x^4 - \cdots)\,dx = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$.

Key identity

From $\dfrac{1}{(1-x)^2} = 1 + 2x + 3x^2 + \cdots$, set $x = \tfrac{1}{2}$: this gives $\displaystyle\sum_{n=1}^{\infty}\dfrac{n}{2^n} = 2$.

§7 — Maclaurin Series from ODEs and Induction AHL 5.19

From first principles (successive differentiation)

1. Compute $f(0), f'(0), f''(0), f'''(0)$.
2. Substitute: $f(x) = f(0) + f'(0) x + \dfrac{f''(0)}{2!} x^2 + \cdots$.
3. Stop at the required degree.

Verify by composition: substitute the series of the inner function and check that coefficients match.

From an ODE (three steps)

Given $\dfrac{dy}{dx} = h(x, y)$ with $y(0) = y_0$:

1. Find $y'(0) = h(0, y_0)$.
2. Differentiate the ODE implicitly to find $y''$; evaluate at $x = 0$.
3. Differentiate again for $y'''$ if needed.
4. Write $y = y_0 + y'(0) x + \dfrac{y''(0)}{2!} x^2 + \cdots$.

Induction formula → Maclaurin coefficients

If $f^{(n)}(x) = Q(n, x) \cdot e^x$ (or another closed form), then $f^{(n)}(0) = Q(n, 0)$, giving

$$f(x) = \sum_{n=0}^{\infty}\frac{Q(n, 0)}{n!} x^n.$$

Example: for $f(x) = x^2 e^x$, induction proves $f^{(n)}(x) = [x^2 + 2nx + n(n-1)] e^x$, so $f^{(n)}(0) = n(n-1)$. Maclaurin: $x^2 e^x = x^2 + x^3 + \dfrac{x^4}{2} + \dfrac{x^5}{6} + \cdots$ (verify by direct expansion ✓).

§8 — Exam Attack Plan All sections

Top 6 marks lost — based on past papers

1. Omitting $\lim_{x \to a}$ notation throughout L'Hôpital working. IB mark schemes explicitly state: no $\lim$ notation $\Rightarrow$ final A1 not awarded. Write $\lim$ on every line until you can substitute.
2. Not stating the indeterminate form before applying L'Hôpital. The R1 mark requires "form $0/0$" or "form $\infty/\infty$" before differentiating. One word, one mark.
3. Stopping a series one term too early. For limit problems, you need the first non-zero term after cancellation. Expand until the $x^n$ you are dividing by cancels cleanly.
4. Wrong validity range for the binomial / log series. Substituting $x \to 3x$ in $\ln(1 + x)$ changes the range $-1 < x \leq 1$ to $-\tfrac{1}{3} < x \leq \tfrac{1}{3}$. Always solve for the new range explicitly.
5. Sign errors in $\ln(1 - x)$ or $(1 - x)^p$. Write $\ln(1 - x) = \ln(1 + (-x))$ and substitute $u = -x$ carefully; every term picks up an extra $(-1)^n$.
6. Trusting the Maclaurin truncation over a large interval. The series for $e^x$, $\sin x$, $\cos x$ converges everywhere, but a finite truncation is only accurate near $x = 0$. For $\int_0^{\pi}$, many terms are needed.

Worked Example — IB-Style Series + Limit

Common Student Questions