L'Hôpital's Rule (pronounced "low-pea-tahl") is a powerful technique for evaluating limits that result in indeterminate forms like 00 or .
📚 Practice Problems
1Problem 1medium
❓ Question:
Evaluate limx→0 using L'Hôpital's Rule.
Explain using:
📋 AP Calculus AB — Exam Format Guide
⏱ 3 hours 15 minutes📝 51 questions📊 4 sections
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💡 Key Test-Day Tips
✓Show all work on FRQs
✓Use proper notation
✓Check units
✓Manage your time
⚠️ Common Mistakes: L'Hôpital's Rule
Avoid these 4 frequent errors
🌍 Real-World Applications: L'Hôpital's Rule
See how this math is used in the real world
📝 Worked Example: Related Rates — Expanding Circle
Problem:
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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L'Hôpital's Rule is part of the AP Calculus AB course on Study Mondo, specifically in the Applications of Derivatives section. You can explore the full course for more related topics and practice resources.
Are there practice problems for L'Hôpital's Rule?
∞
∞
💡 Key Idea: When direct substitution gives 00 or ∞∞, take the derivative of the top and bottom separately, then try again!
The Rule (Formal Statement)
Suppose limx→af(x)=0 and limx→ag(x)=0 (or both approach ±∞).
If limx→ag′(x)f′(x) exists, then:
limx→ag(x)f(x)=limx→ag′(x)f′(x)
Important: This works for x→a, x→a+, x→a−, x→∞, or x→−∞!
When to Use L'Hôpital's Rule
Indeterminate Forms That Work
L'Hôpital's Rule applies ONLY to these forms:
00 ← most common
∞∞ ← also very common
Forms That DON'T Work Directly
These are NOT indeterminate forms for L'Hôpital's Rule:
01 (this is ±∞, not indeterminate)
∞0 (this equals 0)
0∞ (this is ±∞)
But wait! We can sometimes convert other indeterminate forms:
0⋅∞ → rewrite as 00 or ∞∞
∞−∞ → combine and rewrite as a fraction
00, 1∞, ∞0 → use logarithms first!
How to Apply L'Hôpital's Rule
Step-by-Step Process
Step 1: Try direct substitution first
Step 2: Check if you get 00 or ∞∞
If YES → L'Hôpital's Rule applies!
If NO → use another method
Step 3: Take derivatives of numerator and denominator separatelylimx→ag(x)f(x)=limx→ag′(x)f′(x)
Step 4: Evaluate the new limit
If you get a number, you're done!
If you get 00 or ∞∞ again, apply L'Hôpital's Rule again!
If you get something else, stop and use a different method
Example 1: Basic Application
Evaluate limx→0xsinx
Step 1: Direct substitution
0sin0=00 ← indeterminate form!
Step 2: Apply L'Hôpital's Rule
Take derivatives of top and bottom separately:
limx→0xsinx=limx→0(x)′(sinx)′
=limx→01cosx
Step 3: Evaluate
=1cos0=11=1
Answer: limx→0xsinx=1
Example 2: Apply Twice
Evaluate limx→0x21−cosx
First attempt: Direct substitution
021−cos0=01−1=00 ← use L'Hôpital's!
Apply L'Hôpital's Rule (first time)
limx→0x21−cosx=limx→0(x2)′(1−cosx)
=limx→02xsinx
Check: Direct substitution again
2(0)sin0=00 ← still indeterminate!
Apply L'Hôpital's Rule (second time)
limx→02xsinx=limx→0(2x)′(sinx)′
=limx→02cosx
=2cos0=21
Answer: limx→0x21−cosx=21
Example 3: Infinity Form
Evaluate limx→∞xlnx
Step 1: Check the form
As x→∞: lnx→∞ and x→∞
Form: ∞∞ ← L'Hôpital's applies!
Step 2: Apply L'Hôpital's Rule
limx→∞xlnx=limx→∞(x)′(lnx)′
=limx→∞11/x
=limx→∞x1=0
Answer: limx→∞xlnx=0
Interpretation: Logarithms grow slower than linear functions!
Yes, L'Hôpital's would work, but factoring is simpler!
Alternative 2: Standard Limits
You should know: limx→0xsinx=1
Don't need L'Hôpital's every time!
Alternative 3: The Form Doesn't Match
If you get 05, that's ∞ (or −∞), not indeterminate!
Quick Reference Chart
Original Form
What to Do
Result Form
00
L'Hôpital's directly
Apply rule
∞∞
L'Hôpital's directly
Apply rule
0⋅∞
Rewrite as fraction
00 or
∞−∞
Combine fractions
00 or
1∞, 00, ∞0
Take ln first
→ fraction
Growth Rates Using L'Hôpital's
L'Hôpital's Rule helps us compare growth rates:
Logarithms grow slower than polynomials:
limx→∞xplnx=0 for any p>0
Polynomials grow slower than exponentials:
limx→∞exxn=0 for any n
Order from slowest to fastest:
lnx≪xp≪ex≪x!
Historical Note
Named after Guillaume de l'Hôpital (1661-1704), though it was actually discovered by his teacher Johann Bernoulli!
L'Hôpital paid Bernoulli to teach him calculus and share his discoveries. The rule appeared in L'Hôpital's book, the first calculus textbook ever published!
📝 Practice Strategy
Always try direct substitution first - might not be indeterminate!
Check the form - only 00 and ∞∞ work directly
Convert other forms to one of the two usable forms
Take derivatives separately - not the derivative of the quotient!
Simplify when possible - sometimes algebra is faster
Be ready to apply multiple times - but watch for loops
For exponential forms, take ln first, then exponentiate at the end
x2ex−1−x
💡 Show Solution
Step 1: Direct substitution
02e0−1−0=01−1−0=00
Indeterminate form! ✓
Step 2: Apply L'Hôpital's Rule (first time)
Take derivatives of numerator and denominator separately:
Numerator: (ex−1−x)′=ex
Denominator: (x2)′=2x
limx→0x2
Step 3: Check the new limit
2(0)e0−1=
Still indeterminate! Apply L'Hôpital's again.
Step 4: Apply L'Hôpital's Rule (second time)
Numerator: (ex−1)′=ex
Denominator: (2x)′=2
limx→02xe
Step 5: Evaluate
=2e0=2
Answer: limx→0x2
2Problem 2medium
❓ Question:
Evaluate limx→∞exx2 using L'Hôpital's Rule. What does this tell you about growth rates?
💡 Show Solution
Step 1: Check the form
As x→∞: numerator x2→∞ and denominator
3Problem 3expert
❓ Question:
Evaluate limx→0+(sinx)x using L'Hôpital's Rule.
💡 Show Solution
Step 1: Identify the form
As x→0+: sinx→0
4Problem 4medium
❓ Question:
Evaluate lim(x→0) (eˣ - 1)/x using L'Hôpital's Rule.
💡 Show Solution
Step 1: Check form:
As x→0: eˣ - 1 → e⁰ - 1 = 0
x → 0
Form is 0/0 ✓ Can use L'Hôpital's
Step 5: Interpretation:
Logarithm grows slower than any positive power of x
Answer: 0
Mean Value Theorem
▾
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
′
∞∞
∞∞
0⋅∞
−
1
ex−1−x
=
limx→02xex−1
01−1
=
00
x
−
1
=
limx→02ex
1
e
x
−
1
−
x
=
21
ex
→
∞
Form: ∞∞ ← L'Hôpital's applies!
Step 2: Apply L'Hôpital's Rule (first time)
limx→∞exx2=limx→∞(ex)′(x
=limx→∞ex2x
Step 3: Check the form again
As x→∞: 2x→∞ and ex→∞
Still ∞∞ form!
Step 4: Apply L'Hôpital's Rule (second time)
limx→∞ex2x=limx→∞(ex)′(2x)
=limx→∞ex2
Step 5: Evaluate
=∞2=0
Answer: limx→∞exx2=0
Interpretation:
Even though x2 grows to infinity, ex grows so much faster that the ratio goes to 0.
Growth rate conclusion: Exponential functions grow faster than any polynomial!
In general: limx→∞exxn=0 for any positive integer n.
+
So (sinx)x is the form 00 (indeterminate!)
Can't use L'Hôpital's directly on this form.
Step 2: Use logarithms
Let y=(sinx)x
Take natural log of both sides:
lny=ln[(sinx)x]=xln(sinx)