limit(x → π/4) [(sin 2x + cos 3x)/(sin x)] = ....

Jawaban yang benar adalah √(2) – 1 Untuk mencari nilai suatu limit, dengan mensubstitusikan nilai x ke persamaan limitnya. Pada kuadran 2 : cos (π - a) = - cos a Pembahasan : limit(x → π/4) [(sin 2x + cos 3x)/(sin x)] = [(sin 2(π/4) + cos 3(π/4))/(sin (π/4))] = [(sin (2π/4) + cos (3π/4))/(sin (π/4))] = [(sin (π/2) + cos ((4π-π)/4)]/((1/2)√2) = [(1 + cos (4π/4 - π/4)]/((1/2)√2) = [(1 + cos (π - π/4)]/((1/2)√2) = [(1 + (- cos (π/4))]/((1/2)√2) = [(1 – cos (π/4)]/((1/2)√2) = [(1 – (1/2)√2]/((1/2)√2) = (2/2 - √(2)/2)/(√(2)/2) = (2-√(2)/2 · 2/(√(2) = (2-√(2))/(√(2)) = 2/√(2) - √(2)/√(2) = 2√(2)/(√(2) · √(2)) – 1 = 2√(2)/2 – 1 = √(2) – 1 Jadi nilai dari limit(x → π/4) [(sin 2x + cos 3x)/(sin x)] = √(2) – 1