Maths › Further Pure 1 › Limits and L'Hospital's rule
Limits and L'Hospital's rule
When a limit collapses to zero over zero, two rescue routes open. Swap the functions for their series, or swap them for their derivatives. Both read off the answer the fraction was hiding.
Builds on Taylor series and Maclaurin series.
IN THIS TOPIC
- Recognise the indeterminate forms 0/0 and ∞/∞, and say why they carry no verdict.
- Evaluate limits by substituting series expansions and cancelling.
- Apply L'Hospital's rule, repeatedly where needed, and check its conditions first.
- Convert products, differences and power forms into a quotient before starting.
COMMON MISCONCEPTION
If numerator and denominator both tend to zero, the fraction must tend to 1, since they shrink together.
What zero over zero conceals
A quotient whose top and bottom both tend to 0 is an indeterminate form. The outcome depends on how fast each part dies, and the shared destination tells you nothing. The tempting value 1 has no evidence behind it; the true limit can be any number at all, or none. Series make the speeds visible. Replace each function by its expansion and the lowest surviving powers decide the answer.
WORKED EXAMPLE
A limit by series
Find the limit as x → 0 of (x − arctan x)/x³.
arctan x = x − x³/3 + x⁵/5 − …, so the numerator is x³/3 − x⁵/5 + ….
Dividing by x³: 1/3 − x²/5 + … → 1/3.
The x³ terms were the slowest to die on top, and x³ is exactly what the denominator measures.
GUIDED PRACTICE
Another series limit
Find the limit as x → 0 of (e2x² − 1)/x².
Show the working
eu − 1 = u + u²/2 + …, with u = 2x².
The numerator is 2x² + 2x⁴ + …, so dividing by x² gives 2 + 2x² + ….
The limit is 2. One substitution into a standard series and the form stops being indeterminate.
Derivatives to the rescue
L'Hospital's rule says that if f and g are differentiable near the point, though not necessarily at it, with g' non-zero nearby, and both tend to 0, or both to ∞, then lim f/g equals lim f'/g' whenever that second limit exists, finite or infinite. If the new quotient is still indeterminate, apply it again. Each round trades the functions for their rates, and rates are precisely the information that zero over zero was hiding.
Two conditions get skipped in exam scripts and cost marks. Check the form really is indeterminate before differentiating, because the rule gives nonsense on a quotient like sin x/(x + 1). Then differentiate top and bottom separately. The quotient rule has no business here at all.
WORKED EXAMPLE
Three rounds, or one series
Find the limit as x → 0 of (2 sin x − sin 2x)/(x − sin x).
Both parts vanish at 0, and so do their first and second derivatives, so L'Hospital needs three rounds. At the third, the numerator has become −2 cos x + 8 cos 2x and the denominator cos x, giving 6/1.
Series get there in one line. The numerator expands to x³ + … and the denominator to x³/6 + ….
The ratio is x³/(x³/6) → 6. When derivatives keep vanishing, series are usually the faster tool.
Forms that are not quotients yet
Plenty of limits turn up in a form the rule cannot touch. A product of the shape 0 × ∞ becomes 0/0 once you divide by the reciprocal of one factor. A difference of the shape ∞ − ∞ usually yields to a common denominator. Power forms such as 1∞ and 0⁰ need logarithms taken first, and exponentiating at the very end.
Take (1 + a/x)x as x → ∞. Let y be the expression, so ln y = x ln(1 + a/x), and write that as ln(1 + a/x) divided by 1/x. Now the form is 0/0, one round of L'Hospital gives a, and the original limit is ea. That result is worth recognising on sight, since it appears in compound interest questions as often as in pure ones.
ASSESSMENT FOCUS
- Name the indeterminate form before doing anything; the rule only licenses 0/0 and ∞/∞.
- Differentiate top and bottom separately, and never reach for the quotient rule.
- If one round returns 0/0, go again, and state that the form is still indeterminate each time.
- For power forms, take logarithms first and exponentiate at the end. Forgetting to exponentiate is the usual disaster.
- Series are often quicker than three rounds of differentiating; either method earns full marks.
- Quote standard expansions accurately, and keep one term past the one you think you need.
CHECK YOURSELF
Evaluate the limit as x → 0 of sin 3x/x.
Show a hint
One round of L'Hospital, or the first term of the series for sin 3x.
Show the answer
By L'Hospital the quotient becomes 3 cos 3x/1, which tends to 3. Equally, sin 3x = 3x − … so the quotient is 3 − … and again tends to 3.
0/0 says nothing by itself. Expand in series and compare the lowest surviving powers.
L'Hospital replaces f/g by f'/g' while the form stays indeterminate, one named round at a time.
Products, differences and power forms must be forced into a quotient before the rule applies.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the limits and l'hospital's rule questions page.
CHECK YOUR PROGRESS
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- Recognise the indeterminate forms 0/0 and ∞/∞, and say why they carry no verdict.
- Evaluate limits by substituting series expansions and cancelling.
- Apply L'Hospital's rule, repeatedly where needed, and check its conditions first.
- Convert products, differences and power forms into a quotient before starting.
Open the full revision checklist to see every objective in the course in one place.