Does any one know how to multiply Square roots when there also adding a constant?

Question

The problem looks something like this:

Find the limit as x approaches 2 of (6-x)^(1/2) -2 divided by (3-x) ^(1/2) -1.

The problem in words is: the square root of (6-x) minus 2 divided by the square root of (3-x) minus one.

the answer should be 1/2 but i keep getting it wrong. How am i suposed to multiply it?

but why plug in zero?

Answer 1

yes i do

Answer 2

[(6-x)^1/2 -2]/[(3-x)^1/2- 1]
As x --> 2 the expression --> 0/0 so use L'Hospital's rule
[-1/2(6-x)^1/2]/[-1/2(3-x)^1/2]
= -1/2(2)/-1/2 = (-1/4)/-1/2) = 2/4 = 1/2

Answer 3

multiply by the conjugate and then plug in 0