Quick question on limits involving square roots

[banfill_89]

Homework Statement

lim (root*(6-x) -2)/(root*(3-x)-1)
x->2

Homework Equations

i know in a normal limit if a square root was on the top of bottom, you would multiply it and so on so on...but the fact that there is a square root on the top and the bottom is throwing me off.

The Attempt at a Solution

again, i multiplied the top and bottom by root*(6-x) +2/root*(6-x) +2...then that's as far as i get

 

[tiny-tim]

Hi banfill_89!

(have a square-root: √ )

lim (root*(6-x) -2)/(root*(3-x)-1)
x->2
…
again, i multiplied the top and bottom by root*(6-x) +2/root*(6-x) +2...then that's as far as i get

So you got 2 - x on the top, and nasty square-roots on the bottom …

so apply l'Hôpitals' rule!