Nilai x→0lim​ 2xsin 4x1−cos 4x​ = .....

Pertanyaan

Nilai begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application space fraction numerator 1 minus cos invisible function application space 4 x over denominator 2 x sin invisible function application space 4 x end fraction end style = .....

  1. 1

  2. begin mathsize 14px style 1 half end style

  3. 0

  4. begin mathsize 14px style negative 1 half end style

  5. -1

S. Intan

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application space fraction numerator 1 minus cos invisible function application space 4 x over denominator 2 x sin invisible function application space 4 x end fraction  equals limit as x rightwards arrow 0 of invisible function application space fraction numerator 1 minus left parenthesis 1 minus 2 sin squared space invisible function application 2 x right parenthesis over denominator 2 x space open parentheses 2 sin invisible function application space 2 x close parentheses space cos space invisible function application 2 x end fraction  equals limit as x rightwards arrow 0 of invisible function application space fraction numerator 2 sin squared invisible function application space 2 x over denominator 2 x space open parentheses 2 sin invisible function application space 2 x close parentheses space cos space invisible function application 2 x end fraction  equals limit as x rightwards arrow 0 of invisible function application space fraction numerator sin space invisible function application 2 x over denominator 2 x space cos space invisible function application 2 x end fraction  equals limit as x rightwards arrow 0 of invisible function application space fraction numerator tan invisible function application space 2 x over denominator 2 x space end fraction  equals 2 over 2  equals space 1 end style

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Nilai x→0lim​ 2x . sin 2x1−cos x​= .....

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