L'Hopital's Rule
The nuclear option for stubborn limits
You've learned to handle 0/0 by factoring and rationalizing. But what if those techniques don't work? What if you have a limit like:
lim[x->0] sin(x)/x
You can't factor sin(x). You can't rationalize it. This is where L'Hopital's Rule comes in - it's the most powerful technique for handling indeterminate forms.
L'Hopital's Rule
If lim[x->a] f(x)/g(x) = 0/0 or infinity/infinity, then lim[x->a] f(x)/g(x) = lim[x->a] f'(x)/g'(x)
When you get 0/0 or infinity/infinity, take the derivative of the top AND bottom separately, then try the limit again.
⚠️
Critical Warning
L'Hopital's Rule is NOT the quotient rule! You take derivatives of the numerator and denominator separately, not using the quotient rule formula.
Correct: f'(x) / g'(x)
Wrong: (f'g - fg') / g^2
Find lim[x->0] sin(x)/x using L'Hopital's Rule
1
Check the form
Direct substitution: sin(0)/0 = 0/0 ✓ L'Hopital applies!
2
Take derivatives separately
d/dx[sin(x)] = cos(x)
d/dx[x] = 1
3
Apply L'Hopital
lim[x->0] sin(x)/x = lim[x->0] cos(x)/1
4
Evaluate
lim[x->0] cos(x)/1 = cos(0)/1 = 1/1 = 1
Answer:
1
💡 This proves the special trig limit we memorized earlier!
Find lim[x->0] (1 - cos(x))/x^2
1
Check the form
(1 - cos(0))/0^2 = (1-1)/0 = 0/0 ✓
2
First L'Hopital
= lim[x->0] sin(x)/(2x)
3
Check again
sin(0)/(2*0) = 0/0 ✓ Apply L'Hopital again!
4
Second L'Hopital
= lim[x->0] cos(x)/2 = 1/2
Answer:
1/2
1. Find lim[x->0] (e^x - 1)/x using L'Hopital
0/0 form. L'Hopital: lim e^x/1 = e^0/1 = 1
2. Which form allows L'Hopital's Rule?
0/0 only
infinity/infinity only
Both 0/0 and infinity/infinity
Any fraction
L'Hopital works for BOTH 0/0 and infinity/infinity indeterminate forms
Key Takeaways
L'Hopital's Rule: If you get 0/0 or infinity/infinity, differentiate top and bottom separately
You may need to apply it multiple times
NOT the quotient rule - derivatives are taken separately
Only use when you have an indeterminate form!