lim x->2
[((6-x)^1/2)-2]
_______________
[((3-x)^1/2)-1]
Since both numerator and denominator evaluate to zero, we can apply d'Hôpital's rule:
Lim [((6-x)^1/2)-2] / [((3-x)^1/2)-1] x->2
by differentiating numerator and denominator with respect to x.
Lim (-(1/2)/sqrt(6-x)) / (-(1/2)/sqrt(3-x)) as x->2
= Lim sqrt(3-x)/sqrt(6-x) as x->2
= 1/2