The Conjugate Method for Limits

1. Jul 28, 2010

charley076

I'm reviewing limits to tutor a student in precalc and came across a problem.

The conjugate method multiplies the numerator and denominator by the conjugate of the numerator or denominator to simplify the equation. However, after a quick example I wrote for myself, I found that:

lim x-> 3
f(x) / (x-3)
Conjugate Method on the Denominator
lim x-> 3
f(x)*(x+3) / (x^2-9)
The limit still has a 0 in the denominator

lim x-> -3
f(x) / (x+3)
Conjugate Method on the Denominator
lim x-> -3
f(x)*(x-3) / (x^2-9)
The limit still has a 0 in the denominator

The examples of "The Conjugate Method" that I've found online have all multiplied by the conjugate to create a common factor to eliminate the 0 in the denominator.

Ex.
lim x-> 4
(sqrt(x)-2) / x-4
Conjugate of Numerator
lim x-> 4
x-4 / ((x-4)(sqrt(x)+2)
Cancel common factor
lim x-> 4
1 / (sqrt(x)+2)
=1/4

My issue is that the conjugate method is the same as the factoring method as far as I understand it. Is there a difference or is the conjugate method simply easier to use when the factors aren't as easily identifiable.

My point using the example above is:
lim x-> 4
(sqrt(x)-2) / x-4
Factor the denominator into (sqrt(x)-2) & (sqrt(x)+2)
lim x-> 4
(sqrt(x)-2) / ((sqrt(x)-2)(sqrt(x)+2))
Cancel (sqrt(x)-2)
lim x-> 4
1 / (sqrt(x)+2)
= 1/4

Is there a difference between these methods?
Is there a time when only one or the other can be used?

Last edited: Jul 29, 2010
2. Jul 29, 2010

Staff: Mentor

In your first two examples, you have f(x) in the numerator, so multiplying the expression by 1 in the form of the conjugate over itself isn't going to get you anywhere.

The purpose of multiplying by the conjugate over itself is to be able to get common factors in the numerator and denominator that you can eliminate in the hope of being able to evaluate the rest of the limit expression.

3. Jul 29, 2010

charley076

Sorry for the confusion, the reason I put f(x) as the numerator was b/c i didn't know what to put there that wouldn't cause the limit to evaluate to infinity. I just meant f(x) to be filler, i.e. regardless of what the numerator is, multiplying top and bottom by the conjugate of the denominator doesn't resolve the 0 in the denominator.