lim radical 6-x)-2/radical 3-x)-1=
(√(6-x)-2)/(√(3-x)-1)
I assume the limit is as x->2, otherwise there's no problem.
Use l'Hospital's Rule to get
-1/2√(6-x) / -1/2√(3-x)
= √((3-x)/(6-x))
= √(1/4)
= 1/2
I assume the limit is as x->2, otherwise there's no problem.
Use l'Hospital's Rule to get
-1/2√(6-x) / -1/2√(3-x)
= √((3-x)/(6-x))
= √(1/4)
= 1/2