Series & Limits (AA HL)
This topic pulls together two closely linked calculus skills at HL: evaluating limits that don't yield to simple substitution, and expanding a function as an infinite power series using Maclaurin's method. The two connect directly - a Maclaurin series is one of the standard ways to evaluate a tricky limit, alongside l'Hopital's rule. Both are examined without a calculator, so the algebra has to be secure.
What the syllabus says
This topic maps onto three points in the AHL Calculus section of the official IB Analysis & Approaches syllabus.
| Code | Syllabus content |
|---|---|
| AHL5.12 | Informal understanding of continuity and differentiability at a point. Understanding of limits (convergence and divergence). Definition of the derivative from first principles \(f'(x)=\lim\limits_{h\to0}\dfrac{f(x+h)-f(x)}{h}\). Higher derivatives, with notation \(\dfrac{d^ny}{dx^n}\), \(f^{(n)}(x)\). |
| AHL5.13 | Evaluating limits of the form \(\lim\limits_{x\to a}\dfrac{f(x)}{g(x)}\) using l'Hopital's rule or a Maclaurin series, for the indeterminate forms \(\tfrac00\) and \(\tfrac{\infty}{\infty}\). Repeated use of l'Hopital's rule where the first application still gives an indeterminate form. |
| AHL5.19 | Maclaurin series to obtain expansions for \(e^x\), \(\sin x\), \(\cos x\), \(\ln(1+x)\) and \((1+x)^p\) for \(p\in\mathbb{Q}\). Use of simple substitution, products, integration and differentiation to obtain other series. Maclaurin series developed from differential equations. |
These are AHL-only calculus points, examined on Paper 1 (no calculator) and occasionally within longer Paper 2 questions.
Key terms
Five words worth knowing cold before you touch the formulas below - each with a worked example showing exactly what it means.
What is a limit?
A limit describes the value a function approaches as the input gets closer to some point, even if the function isn't actually defined there. Limits let you make sense of expressions like \(\tfrac00\) by looking at what happens nearby rather than exactly at the point.
e.g. \(\lim\limits_{x\to3}\dfrac{x^2-9}{x-3}=\lim\limits_{x\to3}(x+3)=6\).
What is an indeterminate form?
An indeterminate form is an expression like \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) that doesn't have a fixed value on its own - the actual limit depends on how quickly the numerator and denominator approach zero or infinity. Spotting one is the trigger for using l'Hopital's rule or a series.
e.g. \(\lim\limits_{x\to0}\dfrac{\sin x}{x}=1\), even though direct substitution gives \(\tfrac00\).
What is l'Hopital's rule?
L'Hopital's rule says that if \(\lim\limits_{x\to a}\dfrac{f(x)}{g(x)}\) is \(\tfrac00\) or \(\tfrac{\infty}{\infty}\), it equals \(\lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}\) instead - you differentiate the top and bottom separately, not as a quotient. It can be applied more than once if the new limit is still indeterminate.
e.g. \(\lim\limits_{x\to0}\dfrac{e^x-1}{x}=\lim\limits_{x\to0}\dfrac{e^x}{1}=1\).
What is a Maclaurin series?
A Maclaurin series writes a function as an infinite sum of powers of \(x\), built from the function's value and derivatives at \(x=0\). Truncating it after a few terms gives a polynomial approximation that's accurate for \(x\) close to 0.
e.g. \(\cos x\approx1-\dfrac{x^2}{2}\), so \(\cos(0.2)\approx1-0.02=0.98\) (actual value \(0.9801\)).
What is a Taylor series?
A Taylor series is the same idea as a Maclaurin series but centred at any point \(x=a\), not just \(x=0\). It uses the function's value and derivatives at \(x=a\) instead, so a Maclaurin series is just the special case \(a=0\).
e.g. For \(e^x\) about \(x=1\): \(e^x\approx e+e(x-1)\), so \(e^{1.1}\approx e(1+0.1)=1.1e\approx2.990\) (actual value \(3.004\)).
Key formulas
A handful of standard series and one rule for limits cover almost every question on this topic. The two tables below summarise them at a glance - the explanations underneath go into more depth on each one.
Formula reference
The general Maclaurin formula and all five standard expansions are given in the formula booklet - you don't need to memorise them, but you do need to recognise which one to use and how to adapt it.
| Formula | Used for | Booklet? |
|---|---|---|
| \(f(x)=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\cdots\) | General Maclaurin series | ✓ Yes |
| \(e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\) | Maclaurin series for \(e^x\) | ✓ Yes |
| \(\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\) | Maclaurin series for \(\sin x\) | ✓ Yes |
| \(\cos x=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\) | Maclaurin series for \(\cos x\) | ✓ Yes |
| \(\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\cdots\) | Maclaurin series for \(\ln(1+x)\) | ✓ Yes |
| \((1+x)^p=1+px+\dfrac{p(p-1)}{2!}x^2+\cdots\) | Maclaurin series for \((1+x)^p\) | ✓ Yes |
| \(\lim\limits_{x\to a}\dfrac{f(x)}{g(x)}=\lim\limits_{x\to a}\dfrac{f'(x)}{g'(x)}\) | l'Hopital's rule (0/0 or ∞/∞ only) | Not in booklet |
L'Hopital's rule vs the Maclaurin series method
Both techniques evaluate the same indeterminate limits, but they suit different question types - side by side, here's how they compare.
| Feature | L'Hopital's rule | Maclaurin series |
|---|---|---|
| What you do | Differentiate numerator and denominator separately | Replace each function with its power series |
| Best for | A single fraction, one or two derivatives needed | Sums/differences of several functions |
| Repeat if needed? | Yes - reapply while still \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) | Take more terms of the series if the leading terms cancel |
| Example | \(\lim\limits_{x\to0}\dfrac{e^x-1-x}{x^2}\to\dfrac{e^x-1}{2x}\to\dfrac{e^x}{2}=\dfrac12\) | \(\dfrac{(1+x+\tfrac{x^2}{2}+\cdots)-1-x}{x^2}=\dfrac{\tfrac{x^2}{2}+\cdots}{x^2}\to\dfrac12\) |
Standard Maclaurin expansions
These five expansions are the building blocks - every other series on this topic is built from them by substitution, addition or calculus.
\(e^x\) and trig functions
\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\]
\(\sin x\) and \(\cos x\) follow the same pattern, alternating in sign and using only odd (sin) or even (cos) powers.
✓ In the formula booklet\(\ln(1+x)\)
\[\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\]
Alternating signs, no factorials - each term is just \(x^n\) over \(n\).
✓ In the formula booklet\((1+x)^p\)
\[(1+x)^p=1+px+\frac{p(p-1)}{2!}x^2+\cdots\]
Works for any rational \(p\), including negative or fractional values - it's the binomial expansion extended beyond positive integers.
✓ In the formula bookletBuilding new series
Once you know the five standard series, you rarely differentiate a new function from scratch - you adapt a known one.
Substitution
Replace \(x\) in a known series with another expression, such as \(2x\), \(-x\) or \(x^2\), to get the series for a related function.
\(e^{2x}\): replace \(x\) with \(2x\) in the \(e^x\) series.
Adding or subtracting series
Combine two known series term by term - useful for expressions like \(\ln(1+x)-\ln(1-x)\), where the odd-power terms double and the even-power terms cancel.
Always line up matching powers of \(x\) before combining.
Differentiating or integrating
Differentiate or integrate a known series term by term to reach a series for a related function - for instance, integrating the series for \(\tfrac{1}{1+x}\) gives the series for \(\ln(1+x)\).
Only valid within the interval where the original series converges.
Worked examples
Two full exam-style questions, marked exactly like the real thing. Try each one yourself before checking the worked solution.
Consider \(\displaystyle\lim_{x\to2}\dfrac{x^2-x-2}{x-2}.\)
(a) Evaluate this limit.
(b) Deduce, with a reason, whether \(\displaystyle\lim_{x\to2}\dfrac{x^2-x-2}{(x-2)^2}\) exists.
Worked solution
(a) Factorise: \(\dfrac{(x-2)(x+1)}{x-2}\) M1
\(=x+1\) for \(x\neq2.\) A1
As \(x\to2,\ x+1\to3.\) A1
(b) \(\dfrac{x^2-x-2}{(x-2)^2}=\dfrac{x+1}{x-2}\to\infty\) as \(x\to2,\) so the limit does not exist. R1
Consider the Maclaurin series for \(\ln(1-x).\)
(a) Find the series up to and including the term in \(x^4\).
(b) Hence find the Maclaurin series for \(\ln(1+x)-\ln(1-x)\) up to and including the term in \(x^3.\)
Worked solution
(a) Replace \(x\) by \(-x\) in \(\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\tfrac{x^4}{4}+\cdots\) M1
\(\ln(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\cdots\) A1 A1
(b) \(\ln(1+x)-\ln(1-x)=\left(x-\tfrac{x^2}{2}+\tfrac{x^3}{3}\right)-\left(-x-\tfrac{x^2}{2}-\tfrac{x^3}{3}\right)\) M1
\(=2x+\tfrac{2x^3}{3}.\) A1
Common mistakes
The four slip-ups that account for most of the marks lost on this topic - worth reading before you start practising, not just after you get one wrong.
- Treating a Maclaurin series as exact. It's a local approximation valid near \(x=0\) - the further \(x\) is from 0, the worse a truncated series estimates the true value.
- Applying l'Hopital's rule without checking the form first. The rule only applies to \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) - differentiating top and bottom of a limit that isn't indeterminate gives a meaningless answer, even though the steps look identical.
- Dropping or misplacing a factorial. The \(e^x\), \(\sin x\) and \(\cos x\) series divide by \(n!\), not \(n\) - writing \(\tfrac{x^3}{3}\) instead of \(\tfrac{x^3}{3!}=\tfrac{x^3}{6}\) is a very common slip.
- Forgetting the simplified expression only equals the original for \(x\neq a\). After cancelling \((x-a)\) from a limit, the resulting function agrees with the original everywhere except at \(x=a\) itself - which is exactly the point that matters for the limit.
Using your GDC
Limits and Maclaurin series are algebraic Paper 1 skills, so there's no calculator button sequence that replaces the working. Where a limit question does appear on a calculator paper, a graph can help you check your answer, but it should never replace the algebra.
Graph the function near the point in question and zoom in, or use a table of values either side of \(x=a\), to see what value the function seems to be approaching. Most GDCs can also compute a numerical derivative at a point, which can be a useful sanity check if a limit reduces to a derivative-style expression. Be cautious, though - some calculators return a misleading or undefined value exactly at a removable discontinuity (a "hole"), even though the limit itself exists there. Always confirm your final answer algebraically, since these are no-calculator marks on Paper 1.
See the full GDC guide for calculator-specific graphing and table steps.
Ready to practise properly?
Series & limits questions, marked instantly like the real exam.
Quick answers
The questions students on this topic ask most often.
What's the difference between a Maclaurin series and a Taylor series?
A Maclaurin series is a power series expansion of a function centred at \(x=0\). A Taylor series is the same idea centred at any point \(x=a\). Every Maclaurin series is a Taylor series with \(a=0\).
When should I use l'Hopital's rule instead of a Maclaurin series?
Both work on \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) limits. L'Hopital's rule is usually quicker when only one or two derivatives are needed. A Maclaurin series is often clearer when the limit involves several functions added or divided, since you can compare leading terms directly.
Do I have to check the limit is indeterminate before using l'Hopital's rule?
Yes, every time. L'Hopital's rule only applies to the \(\tfrac00\) or \(\tfrac{\infty}{\infty}\) forms. If you differentiate top and bottom when the limit isn't indeterminate, you get the wrong answer even though the method looks the same.
Can I use my GDC for limits and Maclaurin series?
Not directly - these are Paper 1 (no-calculator) skills tested algebraically. On Paper 2 you can graph a function and zoom in near a point to estimate a limit, but you should still confirm the answer algebraically. See the GDC guide for model-specific instructions.
Sub-topics
Series & Limits broken down into its individual skills, each with its own focused page.
Related topics
More Calculus topics from the same AA HL syllabus unit, in case you want to keep going.