Ingat: tan x sin 2x ==cos xsin x2 sin x cos x Maka, diperoleh: limx→0x cos 2xsin x+tan 2x========limx→0x cos 2xsin x+cos 2xsin 2xlimx→0x cos 2xsin x+cos 2x2 sin x cos xlimx→0x cos 2xsin x(1+cos 2x2 cos x)limx→0x sin x+limx→0cos 2x1+cos 2x2 cos x1+11+12×11+11+21+34 Ingat, hasil dari x→0limx cos 2xsin x+tan 2x=4. Jadi, tidak ada pilihan yang tepat.

Hasil dari ....

Pertanyaan

Hasil dari $begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator sin space x plus tan space 2 x over denominator x space cos space 2 x end fraction equals end style$ ....

R. Bella

Master Teacher

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Jawaban terverifikasi

Jawaban

tidak ada pilihan yang tepat.

Pembahasan

Ingat:

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space x space end cell equals cell fraction numerator sin space x over denominator cos space x end fraction end cell row cell sin space 2 x space end cell equals cell 2 space sin space x space cos space x end cell end table$

Maka, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator sin space x plus tan space 2 x over denominator x space cos space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator sin space x plus begin display style fraction numerator sin space 2 x over denominator cos space 2 x end fraction end style over denominator x space cos space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin space x plus begin display style fraction numerator 2 space sin space x space cos space x over denominator cos space 2 x end fraction end style over denominator x space cos space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin space x open parentheses 1 plus begin display style fraction numerator 2 space cos space x over denominator cos space 2 x end fraction end style close parentheses over denominator x space cos space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator sin space x over denominator x space end fraction plus limit as x rightwards arrow 0 of fraction numerator 1 plus fraction numerator 2 space cos space x over denominator cos space 2 x end fraction over denominator cos space 2 x end fraction end cell row blank equals cell 1 plus fraction numerator 1 plus begin display style fraction numerator 2 cross times 1 over denominator 1 end fraction end style over denominator 1 end fraction end cell row blank equals cell 1 plus fraction numerator 1 plus 2 over denominator 1 end fraction end cell row blank equals cell 1 plus 3 end cell row blank equals 4 end table$

Ingat, hasil dari $limit as x rightwards arrow 0 of fraction numerator sin space x plus tan space 2 x over denominator x space cos space 2 x end fraction equals 4$ .

Jadi, tidak ada pilihan yang tepat.