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x → 0 lim ​ sin x cos x x + cos x ​ = .....

.....

  1. 0

  2. 1

  3. 2

  4. 3

  5. 4

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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Pembahasan

begin mathsize 14px style limit as straight x rightwards arrow 0 of invisible function application space fraction numerator straight x plus straight x space cos space invisible function application straight x over denominator sin space invisible function application straight x space cos space invisible function application straight x end fraction  limit as straight x rightwards arrow 0 of invisible function application space fraction numerator straight x open parentheses 1 plus cos space invisible function application straight x close parentheses over denominator sin invisible function application space straight x space cos space invisible function application straight x end fraction  limit as straight x rightwards arrow 0 of invisible function application space fraction numerator 1 plus cos space invisible function application straight x over denominator cos space invisible function application straight x end fraction  equals fraction numerator 1 plus cos space invisible function application 0 over denominator cos space invisible function application 0 end fraction equals fraction numerator 1 plus 1 over denominator 1 end fraction equals 2 end style

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Tentukan nilai limit fungsi berikut. a. x → 8 3 π ​ lim ​ ( sin ( 2 x ) + cos ( 2 x ) 2 cos 2 ( 2 x ) − 2 sin 2 ( 2 x ) ​ )

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