Rationalizing Square Roots
Sometimes you'll encounter limits with square roots that give you 0/0. Factoring won't help here - we need a different technique called rationalizing.
The idea is to multiply by the conjugate to eliminate the square root.
Conjugate
Conjugate
The conjugate of (a + b) is (a - b), and vice versa. When you have a square root, the conjugate changes only the sign between terms.
Classic Rationalization
Medium
Find lim[x->9] (sqrt(x) - 3)/(x - 9)
1
Check direct substitution
(sqrt(9) - 3)/(9 - 9) = (3 - 3)/0 = 0/0 - Indeterminate!
2
Multiply by conjugate/conjugate
(sqrt(x) - 3)/(x - 9) * (sqrt(x) + 3)/(sqrt(x) + 3)
3
Expand the numerator
(sqrt(x) - 3)(sqrt(x) + 3) = x - 9
4
Simplify
(x - 9)/[(x - 9)(sqrt(x) + 3)] = 1/(sqrt(x) + 3)
5
Substitute
lim[x->9] 1/(sqrt(x) + 3) = 1/(3 + 3) = 1/6
Answer:
1/6
Practice Problems
1. Find lim[x->4] (sqrt(x) - 2)/(x - 4)
Multiply by (sqrt(x)+2)/(sqrt(x)+2). Numerator becomes x-4. Cancel to get 1/(sqrt(x)+2). At x=4: 1/4
Key Takeaways
Use rationalization when you have square roots and get 0/0
Multiply by the conjugate over itself (equals 1)
The square root disappears: (sqrt(x)+a)(sqrt(x)-a) = x - a^2