lim (x-> 2) √(ax+b)-3/(x^2-4)=1/4 maka nilai a+b=

Gunakan L'hospital dan ingat jika : y = √f(x) maka y' = f'(x)/2√f(x)

Lim (√(ax + b) - 3)/(x^2 - 4) = 0/0 jika x = 2
(√(a(2) + b) - 3)/(2^2 - 4) = 0/0
√(2a + b) - 3 = 0
√(2a + b) = 3

Lim (√(ax + b) - 3)/(x^2 - 4) = 1/4 jika sesudah diturunkan
x=>2

Lim [a/(2√(ax + b))] / 2x = 1/4
x=>2

[a/(2√(a(2) + b))] / 2(2) = 1/4
[a/2√(2a + b)] / 4 = 1/4
a/2√(2a + b) = 1
a/2(3) = 1
a = 6

√(2a + b) = 3
2a + b = 9
2(6) + b = 9
b = -3

a + b = 6 + (-3) = 3