lim_(x→1/2π) (sinx tan(2x−π)) / (2π−4x)

Jawaban : -1/2 Halo Fania, kakak bantu jawab ya :) Ingat aturan l'Hopital lim_(x→a) f(x)/g(x) = lim_(x→a) f'(x)/g'(x) ingat turunan perkalian (uv)' = u'v + uv' Perhatikan lim_(x→1/2π) (sinx tan(2x−π)) / (2π−4x) f(x) = sinx tan(2x−π) misalkan u = sin x, u' = cos x v = tan (2x-π), v' = 2sec²(2x-π) f'(x) = cos x∙tan (2x-π) + sin x∙2sec²(2x-π) g(x) = (2π−4x) g'(x) = -4 Sehingga lim_(x→1/2π) (sinx tan(2x−π)) / (2π−4x) = lim_(x→1/2π) (sinx tan(2x−π))' / (2π−4x)' = lim_(x→1/2π) (cos x∙tan (2x-π) + sin x∙2sec²(2x-π)) / (-4) = (cos(1/2π)∙tan (2(1/2π)-π) + sin(1/2π)∙2sec²(2(1/2π)-π)) / (-4) = (cos(1/2π)∙tan (0) + sin(1/2π)∙2sec²(0) / (-4) = (0∙0+1∙1)/(-4) = 2/(-4) = -1/2 Jadi, nilai limit tersebut adalah -1/2