limit x -> 1 sin 2(x-1)/(x^2-2x+1)cot 1/2 (x-1)=...

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Pembahasan

Untuk menyelesaikan limit x -> 1 dari [sin(2(x-1))/(x^2-2x+1)] * cot(1/2 (x-1)), kita dapat menggunakan identitas trigonometri dan manipulasi aljabar.

Perhatikan bahwa x^2 - 2x + 1 = (x-1)^2.
Dan cot(theta) = 1/tan(theta).

Limit = limit x -> 1 [sin(2(x-1))/(x-1)^2] * [1/tan(1/2 (x-1))]

Gunakan identitas sin(2a) = 2sin(a)cos(a).
Limit = limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * [1/tan(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [1/tan(1/2 (x-1))]

Kita tahu bahwa limit y -> 0 sin(y)/y = 1. 
Dan limit y -> 0 tan(y)/y = 1.

Mari kita manipulasi agar sesuai dengan bentuk limit tersebut.
Limit = limit x -> 1 [2 * sin(x-1)/(x-1)] * [cos(x-1)] * [ (1/2 (x-1))/tan(1/2 (x-1)) ] * [1 / (1/2 (x-1))]
Limit = limit x -> 1 [2 * sin(x-1)/(x-1)] * [cos(x-1)] * [ (1/2 (x-1))/tan(1/2 (x-1)) ] * [2 / (x-1)]

Saat x -> 1, maka (x-1) -> 0. 
Misalkan y = x-1.
Limit = limit y -> 0 [2 * sin(y)/y] * [cos(y)] * [ (1/2 y)/tan(1/2 y) ] * [2 / y]
Limit = 2 * 1 * cos(0) * (1 / (1/2)) * (2/0) -> ini salah.

Mari kita coba pendekatan lain:
Limit = limit x -> 1 [sin(2(x-1))/(x-1)^2] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]

Kita tahu limit y->0 sin(ky)/y = k. Jadi limit x->1 sin(x-1)/(x-1) = 1.
Kita tahu limit y->0 cos(y) = 1.
Kita tahu limit y->0 sin(y)/y = 1.

Limit = limit x -> 1 [2 * (sin(x-1)/(x-1))] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/ (sin(1/2 (x-1))/(1/2 (x-1))) * (1/2 (x-1))]
Limit = limit x -> 1 [2 * 1] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/(1)] * [1 / (1/2 (x-1))]
Limit = limit x -> 1 2 * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))] * [2 / (x-1)]
Limit = limit x -> 1 4 * cos(x-1) * cos(1/2 (x-1)) / (x-1)^2

Ini masih belum jelas. Mari gunakan L'Hopital's Rule karena bentuknya 0/0.

Bentuk awal: limit x -> 1 sin(2(x-1))/(x^2-2x+1) * cos(1/2 (x-1))/sin(1/2 (x-1))
Bentuknya menjadi: sin(0)/0 * cos(0)/sin(0) = 0/0 * 1/0. Ini bukan bentuk tak tentu 0/0 atau tak hingga/tak hingga.

Periksa kembali soalnya. Mungkin ada kesalahan pengetikan.
Jika soalnya limit x -> 1 [sin(2(x-1))/(x-1)] * [cot(1/2 (x-1)) / (x-1)]

Mari kita asumsikan soalnya adalah limit x -> 1 [sin(2(x-1))/(x-1)] * [cot(1/2 (x-1))]

Limit = limit x -> 1 [sin(2(x-1))/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)/(x-1)] * cos(x-1) * [cos(1/2 (x-1))/sin(1/2 (x-1))]

Kita tahu limit y->0 sin(ky)/y = k dan limit y->0 tan(ky)/y = k.
Limit = limit x -> 1 [2 * 1] * cos(0) * [cos(1/2 (x-1))/(sin(1/2 (x-1))/(1/2 (x-1))) * (1/2 (x-1))]
Limit = 2 * 1 * [1 / (1)] * [1 / (1/2 * (x-1))]
Limit = 2 * [1 / (1/2 * (x-1))]
Limit = 4 / (x-1) -> tak hingga.

Mari kita coba interpretasi lain dari soal:
limit x -> 1 [sin(2(x-1))/(x^2-2x+1)] * cot(1/2 (x-1))
= limit x -> 1 [sin(2(x-1))/(x-1)^2] * cot(1/2 (x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * cos(1/2 (x-1))/sin(1/2 (x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]

Kita gunakan sin(A) = 2sin(A/2)cos(A/2).
Misal A = x-1.
sin(x-1) = 2sin((x-1)/2)cos((x-1)/2).

Limit = limit x -> 1 [2 * 2sin((x-1)/2)cos((x-1)/2) / (x-1)] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [4sin((x-1)/2)cos((x-1)/2) / (2 * (x-1)/2)] * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin((x-1)/2)/((x-1)/2)] * cos((x-1)/2) * [cos(x-1)/(x-1)] * [cos(1/2 (x-1))/sin(1/2 (x-1))]

Saat x -> 1:
[2sin((x-1)/2)/((x-1)/2)] -> 2 * 1 = 2
cos((x-1)/2) -> cos(0) = 1
cos(x-1) -> cos(0) = 1
(x-1) -> 0
cos(1/2 (x-1)) -> cos(0) = 1
sin(1/2 (x-1)) -> sin(0) = 0

Bentuknya menjadi 2 * 1 * [1/0] * [1/0] -> tak hingga.

Kemungkinan besar ada kesalahan pada soal. Jika kita asumsikan soalnya adalah:
limit x -> 0 sin(2x)/(x^2) cot(x/2)
= limit x -> 0 sin(2x)/x^2 * cos(x/2)/sin(x/2)
= limit x -> 0 [2sin(x)cos(x)/x^2] * [cos(x/2)/sin(x/2)]
= limit x -> 0 [2sin(x)/x * cos(x)/x] * [cos(x/2)/sin(x/2)]

Jika kita kembali ke soal asli dan menggunakan L'Hopital:
Limit = limit x -> 1 sin(2(x-1))/(x^2-2x+1)cot(1/2 (x-1))
Bentuk 0/0 jika kita tulis sin(2(x-1))/(x-1)^2 * cos(1/2(x-1))/sin(1/2(x-1))
Turunan pembilang: 2cos(2(x-1)) * cot(1/2(x-1)) + sin(2(x-1)) * (-csc^2(1/2(x-1)) * 1/2)
Turunan penyebut: 2(x-1)

Saat x=1, turunan pembilang = 2cos(0)*cot(0) + sin(0)*... -> tak hingga.
Turunan penyebut = 0.
Ini menunjukkan bahwa limitnya tidak terhingga atau tidak terdefinisi.

Mari kita ubah bentuknya:
Limit = limit x -> 1 [sin(2(x-1)) / (x-1)] * [1 / (x-1)] * [cos(1/2 (x-1)) / sin(1/2 (x-1))]
Limit = limit x -> 1 [2sin(x-1)cos(x-1) / (x-1)] * [cos(1/2 (x-1)) / sin(1/2 (x-1))] * [1 / (x-1)]
Limit = limit x -> 1 [2sin(x-1)/(x-1)] * cos(x-1) * [cos(1/2 (x-1)) / sin(1/2 (x-1))] * [1 / (x-1)]
Limit = 2 * 1 * 1 * [cos(1/2 (x-1)) / sin(1/2 (x-1))] * [1 / (x-1)]
Limit = 2 * [cos(1/2 (x-1)) / sin(1/2 (x-1))] * [1 / (x-1)]
Limit = 2 * cos(1/2 (x-1)) / [ (x-1)sin(1/2 (x-1)) ]
Limit = 2 * cos(1/2 (x-1)) / [ 2 * (x-1)/2 * sin(1/2 (x-1)) ]
Limit = cos(1/2 (x-1)) / [ (x-1)/2 * sin(1/2 (x-1)) ]
Limit = cos(1/2 (x-1)) / [ (x-1)/2 ] * [ 1 / sin(1/2 (x-1)) ]
Limit = cos(1/2 (x-1)) * [ 1 / ((x-1)/2) ] * [ (1/2 (x-1)) / sin(1/2 (x-1)) ] * [ 1 / (1/2 (x-1)) ]
Limit = cos(1/2 (x-1)) * [ 2/(x-1) ] * [ (1/2 (x-1)) / sin(1/2 (x-1)) ] * [ 2/(x-1) ]

Mari kita coba manipulasi lain:
Limit = limit x -> 1 [sin(2(x-1))/(x-1)^2] * cot(1/2 (x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * cos(1/2(x-1))/sin(1/2(x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [cos(1/2(x-1))/sin(1/2(x-1))]
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [1/tan(1/2(x-1))]

Kita tahu limit y->0 sin(ky)/y = k dan limit y->0 tan(ky)/y = k.
Limit = limit x -> 1 [2 * 1] * [cos(x-1)/(x-1)] * [1 / ( (1/2)(x-1) * tan(1/2(x-1)) / (1/2)(x-1) ) ]
Limit = 2 * [cos(x-1)/(x-1)] * [1 / ( (1/2)(x-1) * 1 ) ]
Limit = 2 * [cos(x-1)/(x-1)] * [2 / (x-1)]
Limit = 4 * cos(x-1) / (x-1)^2
Saat x->1, cos(x-1)->1 dan (x-1)^2->0. Maka limitnya adalah tak hingga.

Asumsikan soalnya adalah limit x -> 1 [sin(2(x-1))/(x-1)] * cot(1/2 (x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)] * cos(1/2(x-1))/sin(1/2(x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * cos(x-1) * [cos(1/2(x-1))/sin(1/2(x-1))]
= 2 * 1 * 1 * [cos(1/2(x-1)) / ( (1/2)(x-1) * sin(1/2(x-1))/(1/2)(x-1) )]
= 2 * [cos(1/2(x-1)) / ( (1/2)(x-1) * 1 )]
= 4 * cos(1/2(x-1)) / (x-1)
Ini juga tak hingga.

Jika kita mengasumsikan soalnya adalah limit x -> 0 sin(2x)/x * cot(x/2)
= limit x -> 0 sin(2x)/x * cos(x/2)/sin(x/2)
= limit x -> 0 [2sin(x)cos(x)/x] * [cos(x/2)/sin(x/2)]
= limit x -> 0 [2sin(x)/x] * cos(x) * [cos(x/2)/sin(x/2)]
= 2 * 1 * 1 * [cos(x/2) / ( (x/2) * sin(x/2)/(x/2) )]
= 2 * [cos(x/2) / ( (x/2) * 1 )]
= 4 * cos(x/2) / x
Ini juga tak hingga.

Asumsikan soalnya adalah limit x -> 1 sin(2(x-1))/((x-1)^2) * tan(1/2(x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * tan(1/2(x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * tan(1/2(x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [ (1/2)(x-1) * tan(1/2(x-1))/(1/2)(x-1) ]
= limit x -> 1 [2 * 1] * [cos(x-1)/(x-1)] * [ (1/2)(x-1) * 1 ]
= limit x -> 1 2 * [cos(x-1)/(x-1)] * (1/2)(x-1)
= limit x -> 1 sin(x-1) * cos(x-1)
= sin(0) * cos(0) = 0 * 1 = 0.

Jika soalnya adalah:
limit x -> 1 sin(2(x-1))/(x-1) * cot(1/2 (x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)] * cos(1/2(x-1))/sin(1/2(x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * cos(x-1) * [cos(1/2(x-1))/sin(1/2(x-1))]
= 2 * 1 * 1 * [cos(1/2(x-1)) / ( (1/2)(x-1) * sin(1/2(x-1))/(1/2)(x-1) )]
= 2 * [cos(1/2(x-1)) / ( (1/2)(x-1) * 1 )]
= 4 * cos(1/2(x-1)) / (x-1)
Ini tak hingga.

Jawaban yang paling mungkin didapatkan jika soalnya adalah:
limit x -> 1 [sin(2(x-1))/(x-1)] * [tan(1/2 (x-1))/(x-1)]
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)] * [tan(1/2 (x-1))/(x-1)]
= limit x -> 1 [2sin(x-1)/(x-1)] * cos(x-1) * [tan(1/2 (x-1))/(x-1)]
= 2 * 1 * 1 * [ (1/2)(x-1) * tan(1/2 (x-1))/(1/2)(x-1) ] / (x-1)
= 2 * [ (1/2)(x-1) * 1 ] / (x-1)
= 2 * (1/2)(x-1) / (x-1) = 1.

Jika soalnya adalah limit x -> 1 [sin(2(x-1))/(x^2-2x+1)] * tan(1/2 (x-1))
= limit x -> 1 [2sin(x-1)cos(x-1)/(x-1)^2] * tan(1/2 (x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * tan(1/2 (x-1))
= limit x -> 1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [ (1/2)(x-1) * tan(1/2 (x-1))/(1/2)(x-1) ]
= 2 * 1 * [cos(x-1)/(x-1)] * [ (1/2)(x-1) * 1 ]
= 2 * [cos(x-1)/(x-1)] * (1/2)(x-1)
= sin(x-1)cos(x-1) = 0.

Mari kita anggap soalnya adalah mencari nilai:
limit x->1 sin[2(x-1)] / [(x^2-2x+1)cot(1/2(x-1))]
= limit x->1 sin[2(x-1)] / [(x-1)^2 cot(1/2(x-1))]
= limit x->1 [2sin(x-1)cos(x-1) / (x-1)^2] * tan(1/2(x-1))
= limit x->1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * tan(1/2(x-1))
= limit x->1 [2sin(x-1)/(x-1)] * [cos(x-1)/(x-1)] * [(1/2)(x-1) * tan(1/2(x-1))/(1/2(x-1))]
= 2 * 1 * [cos(x-1)/(x-1)] * (1/2)(x-1) * 1
= 2 * cos(x-1) * (1/2) = cos(x-1)
Saat x->1, cos(x-1) -> cos(0) = 1.

Jadi, jika soalnya seperti yang tertulis, jawabannya adalah 1.