When to Use L'Hopital's Rule
L'Hopital's Rule is powerful, but it's not always the best choice. Sometimes factoring is faster. Sometimes L'Hopital makes things worse. Let's learn when to use it - and when NOT to.
Decision Framework
Use L'Hopital when:
- Factoring doesn't work (trig, exponentials, logs)
- The expression is complicated
- You've verified it's 0/0 or infinity/infinity
Don't use L'Hopital when:
- Simple factoring will work
- It's NOT an indeterminate form
- The derivatives make things more complex
When NOT to Use L'Hopital
Easy
Find lim[x->2] (x^2 - 4)/(x - 2)
1
Bad approach: L'Hopital
Would give: 2x/1 = 4 at x=2
This works but is overkill!
2
Better approach: Factor
(x^2-4)/(x-2) = (x+2)(x-2)/(x-2) = x+2 = 4
Answer:
4
💡 Both methods work, but factoring is simpler for polynomials!
Infinity/Infinity Form
Medium
Find lim[x->infinity] (3x^2 + 2x)/(5x^2 - 1)
1
Check the form
As x->infinity, both top and bottom go to infinity: infinity/infinity ✓
2
Apply L'Hopital
= lim[x->infinity] (6x + 2)/(10x)
Still infinity/infinity!
3
Apply L'Hopital again
= lim[x->infinity] 6/10 = 3/5
Answer:
3/5
💡 For rational functions, the limit equals the ratio of leading coefficients when degrees are equal
Practice
1. lim[x->infinity] (2x^3)/(x^3 + 1) = ?
Infinity/infinity. L'Hopital twice gives 12x/6x = 2. Or: leading coefficients 2/1 = 2
Key Takeaways
Check if you have 0/0 or infinity/infinity FIRST
Try simple methods (factoring) before L'Hopital
L'Hopital shines with trig, exponentials, and logs
You can apply L'Hopital multiple times if needed