Evaluate: (without using Hopital's rule)
d) lim (6-3x)/(((5x+6)^1/2)-4)
x->2
(6-3x)/[sqrt(5x+6)-4] =
(6-3x)[sqrt(5x+6)+4]/{[sqrt(5x+6)+4][sqrt(5x+6)-4]} =
(6-3x)[sqrt(5x+6)+4]/(5x+6 - 16) =
(6-3x)[sqrt(5x+6)+4]/(5x-10)
The limit of x to 2 of
(6-3x)/(5x-10)
can be computed by expressing e.g. the denominator in terms of the numerator:
5 x - 10 = -5/3 (6-3x)
So, that limit is -3/5 and multiplying with the value of sqrt(5x+6)+4 at x = 2 gives
-5/3 * 8 = -40/3
for the limit of the original function.