Answer:
Option C
Explanation:
We have
L = limx→2√(x+7)−3√(2x−3)3√(x+6)−23√(3x−5)
(00 form)
Let x-2 = t such that when x→2, t→0. Then
L= limt→0(t+9)1/2−3(2t+1)1/2(t+8)1/3−2(3t+1)1/3
= 32limt→0(1+t/9)1/2−(2t+1)1/2(1+t/8)1/3−(3t+1)1/3
= 32limt→0(12t9)−(2t)1213t8−(3t)13
= 32118−1124−1
= 3423