# Evaluate the limit of [(6-x)^0.5 - 2]/[(3-x)^0.5 - 1]

MathewsMD
Evaluate:

lim [(6-x)^0.5 - 2]/[(3-x)^0.5 - 1]
x→2

I have tried rationalizing, but doing so does not help.

I have also tried using the squeeze theorem to solve the question, but I have not found any values that can be used to precisely find the limit. The values I could think of were -x^2 and and x^2, but that does not allow you to find the exact answer.

Any help or tips with this question would be greatly appreciated!

Have you tried L'Hospital's rule?

MathewsMD
I just started introductory calculus and have not learned it yet. Any help with or without using L'Hospital's rule would be great!

Gustafo
I am pretty sure that it is going to be impossible for you to evaluate this limit analytically without the use of L'Hospital's rule. May I ask exactly what level of math you are in and what chapter? I am thinking they might want you to do this using the numerical approach. Which is to say the least, very tedious.

iRaid
Multiply the top and bottom by (1/(3-x)^.5) I think that'll do it.

MathewsMD
I am currently on Chapter 2 in Stewart's Calculus (Calculus I). I've learned Squeeze Theorem and other basic limit solving strategies, but nothing too advanced.

And I believe I have tried rationalizing using that root, but have not gotten a solution.

Homework Helper
This may be a big hint, but you will have to rationalize twice. First, rationalize the numerator. Simplify the numerator, but don't multiply it out in the denominator. Then rationalize the denominator. Here, you will need to multiply out two of the factors in the denominator. Eventually, you'll be able to apply direct substitution.

phrox
By rationalize do you mean multiply the conjugate? I don't see how you can rationalize (multiply top and bottom by the root), but you say you only do it to the top and then only do it to the bottom. I don't think you can do that. Also if I multiply the top conjugate and then bottom conjugate it gets REALLY messy and it can't be right.

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