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Question

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L. Jannah · Accepted Answer

Jawaban: 1/24

y = sin x → y' = cos x

y = cos ax → y' = - a sin ax

y = tan ax → y' = a sec² ax

y = x → y' = 1

y = (cos ax)^n → y' = n·a·(- sin ax)·(cos ax)^[n-1]

lim (x → a) f(x)·g(x) = lim (x → a) f(x) · lim (x → a) g(x)

lim (x → 0)(sin ax/sin bx) = a/b

sec x = 1/cos x

Aturan L'Hospital

lim (x → a) f(x)/g(x) = lim (x → a) f'(x)/g'(x)

Langkah pertama menentukan limit, gunakan aturan substitusi langsung.

lim (x → 0) (x - sin x)/(tan 2x - sin 2x)

= (0 - sin 0)/(tan 2(0) - sin 2(0))

= 0/0

karena hasilnya tak tentu (0/0), maka gunakan cara lain.

dengan Aturan L'Hospital, maka:

lim (x → 0) (x - sin x)/(tan 2x - sin 2x)

= lim (x → 0) (1 - cos x)/(2 sec² 2x - 2 cos 2x)

= lim (x → 0) (1 - cos x)/(2 (sec² 2x - cos 2x))

= ½ lim (x → 0) (1 - cos x)/(sec² 2x - cos 2x)

= ½ lim (x → 0) (1 - cos x)/(1/(cos² 2x) - cos 2x)

= ½ lim (x → 0) (1 - cos x)/[(1 - cos³ 2x)/cos² 2x]

= ½ lim (x → 0) (1 - cos x)·[cos² 2x/(1 - cos³ 2x)]

= ½ lim (x → 0) [cos² 2x(1 - cos x)]/(1 - cos³ 2x)

= ½ lim (x → 0) cos² 2x · lim (x → 0)(1 - cos x)/(1 - cos³ 2x)

= ½ · cos² 2(0) · lim (x → 0)(1 - cos x)/(1 - cos³ 2x)

= ½ · 1² · lim (x → 0)(1 - cos x)/(1 - cos³ 2x)

= ½ lim (x → 0)(1 - cos x)/(1 - cos³ 2x)

= ½ lim (x → 0)(cos x - 1)/(cos³ 2x - 1)

Gunakan aturan L'Hospital kembali karena nilai limitnya masih 0/0.

Maka,

½ lim (x → 0)(cos x - 1)/(cos³ 2x - 1)

= ½ lim (x → 0)(- sin x - 0)/(3·2 cos² 2x (- sin 2x) - 0)

= ½ lim (x → 0)(- sin x )/(-6 sin 2x cos² 2x)

= ½ lim (x → 0)(sin x )/(6 sin 2x cos² 2x)

= ½·(1/6) · lim (x → 0)(sin x )/(sin 2x cos² 2x)

= (1/12) · lim (x → 0)(1/cos² 2x) · lim (x → 0)(sin x/sin 2x)

= (1/12) · (1/1) · lim (x → 0)(sin x/sin 2x)

= (1/12) · 1 · ½

= 1/24

Jadi, nilai dari limit tersebut adalah 1/24.

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