Hasil x→0lim​cos2 2x−1x tan 4x​ adalah ....

Pertanyaan

Hasil limit as x rightwards arrow 0 of fraction numerator x space tan space 4 x over denominator cos squared space 2 x minus 1 end fraction adalah ....

  1. negative 4

  2. negative 2 

  3. negative 1

  4. 1

  5. 2

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Rumus limit trigonometri untuk x rightwards arrow 0 yang didapat dari penurunan dengan kaidah L'Hopital sebagai berikut.

left parenthesis straight i right parenthesis space limit as x rightwards arrow 0 of fraction numerator a x over denominator sin space b x end fraction equals a over b left parenthesis ii right parenthesis space limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b

Selain itu, ingatlah identitas trigonometri berikut.

left parenthesis straight i right parenthesis space tan space x equals fraction numerator sin space x over denominator cos space x end fraction left parenthesis ii right parenthesis space sin squared open parentheses a x close parentheses plus cos squared open parentheses a x close parentheses equals 1

Pertama, lakukan metode substitusi untuk menjawab soal limit di atas.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator x space tan space 4 x over denominator cos squared space 2 x minus 1 end fraction end cell equals cell fraction numerator 0 space tan space 4 left parenthesis 0 right parenthesis over denominator left parenthesis cos space 2 left parenthesis 0 right parenthesis right parenthesis squared minus 1 end fraction end cell row blank equals cell fraction numerator 0 times 0 over denominator 1 minus 1 end fraction end cell row blank equals cell 0 over 0 space left parenthesis bentuk space tak space tentu right parenthesis end cell end table

Karena menghasilkan bentuk tak tentu, penyelesaian limit di atas adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator x space tan space 4 x over denominator cos squared space 2 x minus 1 end fraction end cell equals cell limit as x rightwards arrow 0 of open parentheses x times fraction numerator sin space 4 x over denominator cos space 4 x end fraction times fraction numerator 1 over denominator cos squared space 2 x minus open parentheses sin squared space 2 x plus cos squared space 2 x close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses x times fraction numerator sin space 4 x over denominator cos space 4 x end fraction times fraction numerator 1 over denominator negative sin squared space 2 x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses negative fraction numerator x over denominator sin space 2 x end fraction times fraction numerator sin space 4 x over denominator sin space 2 x end fraction times fraction numerator 1 over denominator cos space 4 x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses negative fraction numerator x over denominator sin space 2 x end fraction close parentheses times limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 4 x over denominator sin space 2 x end fraction close parentheses times limit as x rightwards arrow 0 of open parentheses fraction numerator 1 over denominator cos space 4 x end fraction close parentheses end cell row blank equals cell negative 1 half times 4 over 2 times fraction numerator 1 over denominator cos space 4 left parenthesis 0 right parenthesis end fraction end cell row blank equals cell negative 1 half times 2 times 1 over 1 end cell row blank equals cell negative 1 end cell end table

Dengan demikian, hasil dari limit as x rightwards arrow 0 of fraction numerator x space tan space 4 x over denominator cos squared space 2 x minus 1 end fraction adalah negative 1.

Oleh karena itu, jawaban yang benar adalah C.

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