SOLUTION: This is actually a Calculus 1 problem, but I am having problems with the algebra portion. So far, I have determined that I need to multiply this problem by the conjugate of BOTH th

Question 989465: This is actually a Calculus 1 problem, but I am having problems with the algebra portion. So far, I have determined that I need to multiply this problem by the conjugate of BOTH the numerator and denominator, but I've never done that. Once I get the fraction all worked out and cleaned up, I know how to handle the limit portion.
Evaluate the limit of:
(((6-x)^(1/2))-2) / (((3-x)^(1/2))-1)
as x approaches 2
Here is my work:
(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1) * (((6-x)^(1/2))-2)/(((6-x)^(1/2))-2)
=> (((6-x)^2)-((4)^2))/(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1)
Then do I multiply out the top, and then multiply by the conjugate of the numerator?
=> ((-x^2)-12x+36)/(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1) * (((3-x)^(1/2))-1)/(((3-x)^(1/2))-1)
This is where I'm really confused. How do I multiply a polynomial by the conjugate? Would the numerator be ((-x^2)(3-(x^(1/2)))-12x((3-(x^(1/2))+36((3-x)^(1/2))-x^2+12x-36?
Now what? I appreciate any and all help. Please let me know where I went wrong, and what I need to do to work out the problem.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

Well, to answer your question straight-up, The conjugate of a binomial expression is simply the same expression with the sign in the middle changed. And the product of a binomial and its conjugate is the difference of two squares.

What it looks like you are trying to do is to rationalize both your numerator and denominator; maybe you have an irrational (no pun intended) fear of radicals. If that is the case, I hate to be the bearer of bad tidings, but you are doomed to failure. Yes, you can rationalize your numerator, at the expense of ending up with an EUPC (that's Extremely Ugly Piece of Crap) in the denominator. OR you can rationalize your denominator, at the expense of the mathematical equivalent of pornography in the numerator. But not both. And even at that, once you have either a rational numerator or denominator, that rational expression is still going to be zero in the limit, and its ugly companion will also be zero in the limit. Life lesson here: Once you have gotten yourself into a hole, stop digging.

However, the larger question is: Why are you mucking about with multiplying by conjugates? What you have is a L'H�pital indeterminate form