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lim
x->~ (ax+b) - V(a^2 x^2 + px + q)= (2ab-p)/2a
lim (3x - 2) - V(9x^2 - 2x + 5)
x-> ~
= (2(3)(-2) - (-2))/2(3)
= (-10)/6
= -5/3
= -1 1/3
2) Karena disubstitusikan hasilnya 0/0 maka pakai dalil L'Hospital ..
Semoga membantu :)
lim x → ∞
(3x - 2) - √(9x² - 2x + 5)
= √(3x - 2)² - √(9x² - 2x + 5)
= √(9x² - 12x + 4) - √(9x² - 2x + 5)
Ingat sifat ini :
lim x →
∞
√(ax² + bx + c) - √(px² + qx + r)
= (b - q) / 2√a
dengan syarat a = p
Maka :
lim x → ∞
√(9x² - 12x + 4) - √(9x² - 2x + 5)
= (-12 - (-2) / 2√9
= (-12 + 2)/ 2(3)
= -10/6
= -5/3
= -2 ¹/₃
Jadi, nilai limitnya adalah -2 ¹/₃ (Jawaban B)
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11.)
lim x → 0
x³ - 9x
{ _______ }³
x² + 3x
x(x² - 9)
= { _______ }³
x(x + 3)
x(x + 3) (x - 3)
= { ____________}³
x(x + 3)
= (x - 3)³
= (0 - 3)³
= (-3)³
= -27
Jadi, nilai limitnya adalah -27 (Jawaban A)