Nilai dari lim_(f→0) (1−√(1+f))/(f^(2)−f)=⋯

Jawaban yang benar adalah 1/2. Pembahasan : Konsep : (a−√b)(a+√b) = a² − b lim_(f→0) (1−√(1+f))/(f²−f) = (1−√(1+0))/(0²−0) = (1−√1)/(0−0) = (1−1)/0 = 0/0 (tak tentu) Karena hasilnya 0/0 maka gunakan perkalian sekawan. lim_(f→0) (1−√(1+f))/(f²−f) = lim_(f→0) (1−√(1+f))/(f²−f) . (1+√(1+f))/(1+√(1+f)) = lim_(f→0) (1−(1+f)) / (f²−f)(1+√(1+f)) = lim_(f→0) (1−1−f) / f(f−1)(1+√(1+f)) = lim_(f→0) (−f) / f(f−1)(1+√(1+f)) = lim_(f→0) −1 / (f−1)(1+√(1+f)) = −1 / (0−1)(1+√(1+0)) = −1 / (−1)(1+√1) = −1 / (−1)(1+1) = −1 / (−1)(2) = 1/2 Jadi, lim_(f→0) (1−√(1+f))/(f²−f) = 1/2 Semoga membantu ya.