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 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 why they carry no verdict.
- Evaluate limits by substituting series expansions and cancelling.
- Apply L'Hospital's rule, repeatedly where needed, and know when series are quicker.
WHAT YOU PROBABLY THINK
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, not on the fact that both do. The opening claim picks 1 with no evidence; the true limit can be anything. Series make the speeds visible: replace each function by its expansion and the lowest surviving powers decide.
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 they are exactly what the denominator measures.
YOUR TURN
Another series limit
Find the limit as x → 0 of (e^(2x²) − 1)/x².
Show the working
e^(u) − 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: if f and g both tend to 0 (or both to ∞), then lim f/g = lim f'/g', provided the second limit exists. If the new quotient is still 0/0, apply it again. Each round trades the functions for their rates, which is exactly the information zero over zero was hiding.
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: L'Hospital needs three rounds, ending with (−2 cos x + 8 cos 2x)/... at the third differentiation, giving 6.
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 like (1 + a/x)^x as x → ∞ convert first: take logarithms, rewrite as a fraction, and the rule applies. The answer e^(a) drops out, a limit worth recognising on sight.
THE EXAM BIT
- Name the indeterminate form before doing anything; the rule only applies to 0/0 and ∞/∞.
- Differentiate top and bottom separately; the quotient rule has no business here.
- If one round of L'Hospital returns 0/0, go again, and say so each time.
- For power forms, take logs first and exponentiate at the end.
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
B
y
L
'
H
o
s
p
i
t
a
l
:
3
c
o
s
3
x
/
1
→
3
.
E
q
u
a
l
l
y
,
s
i
n
3
x
=
3
x
−
…
s
o
t
h
e
q
u
o
t
i
e
n
t
i
s
3
−
…
→
3
.
0/0 says nothing by itself: expand in series and compare the lowest surviving powers.
L'Hospital: replace f/g by f'/g' while the form stays indeterminate, one named round at a time.
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.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- Recognise the indeterminate forms 0/0 and ∞/∞ and why they carry no verdict.
- Evaluate limits by substituting series expansions and cancelling.
- Apply L'Hospital's rule, repeatedly where needed, and know when series are quicker.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.