x→0lim sin223x (1−sec22x)sin43x=...

Pertanyaan

$limit as x rightwards arrow 0 of space fraction numerator sin to the power of 4 3 x over denominator sin squared begin display style 3 over 2 end style x space left parenthesis 1 minus sec squared 2 x right parenthesis end fraction equals...$

$9$
$\frac94$
$\frac9{16}$
$-\frac94$
$-9$

W. Rohmiyati

Master Teacher

Mahasiswa/Alumni Universitas Langlangbuana

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Untuk mengerjakan soal limit fungsi trigonometri, kita siubtitusi terlebih dahulu menjadi:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator sin to the power of 4 3 x over denominator sin squared begin display style 3 over 2 end style x space left parenthesis 1 minus sec squared 2 x right parenthesis end fraction end cell equals cell fraction numerator sin to the power of 4 3 open parentheses 0 close parentheses over denominator sin squared begin display style 3 over 2 end style left parenthesis 0 right parenthesis space left parenthesis 1 minus sec squared 2 left parenthesis 0 right parenthesis end fraction end cell row blank equals cell fraction numerator sin to the power of 4 left parenthesis 0 right parenthesis over denominator sin squared left parenthesis 0 right parenthesis space left parenthesis 1 minus sec squared left parenthesis 0 right parenthesis right parenthesis end fraction end cell row blank equals cell fraction numerator left parenthesis sin space 0 right parenthesis to the power of 4 over denominator left parenthesis sin space 0 right parenthesis squared space left parenthesis 1 minus left parenthesis sec space 0 right parenthesis squared right parenthesis end fraction end cell row blank equals cell fraction numerator 0 to the power of 4 over denominator 0 squared left parenthesis 1 minus 1 squared right parenthesis end fraction end cell row blank equals cell fraction numerator 0 over denominator 0 left parenthesis 0 right parenthesis end fraction end cell row blank equals cell 0 over 0 space space left parenthesis Nilai space tak space tentu right parenthesis end cell end table$

Karena diperoleh bentuk tak tentu, maka limit fungsi trigonometri diubah lalu gunakan konsep

$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b limit as x rightwards arrow 0 of space fraction numerator sin space a x over denominator tan space b x end fraction equals a over b$

serta identitas trigonometri berikut ini

$\begin{array}{rcl}\tan^2x+1&=&\sec^2x\\1-\sec^2x&=&-\tan^2x\end{array}$

Setelah itu, maka soal di atas diselesaikan dengan cara berikut ini:

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of space fraction numerator sin to the power of 4 3 x over denominator sin squared begin display style 3 over 2 end style x space left parenthesis 1 minus sec squared 2 x right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator sin 3 x times sin 3 x times sin 3 x times sin 3 x over denominator sin begin display style 3 over 2 end style x times sin begin display style 3 over 2 end style x space left parenthesis negative tan squared 2 x right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator sin 3 x times sin 3 x times sin 3 x times sin 3 x over denominator sin begin display style 3 over 2 end style x times sin begin display style 3 over 2 end style x times negative tan 2 x times tan 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator up diagonal strike sin 3 up diagonal strike x times up diagonal strike sin 3 up diagonal strike x times up diagonal strike sin 3 up diagonal strike x times up diagonal strike sin 3 up diagonal strike x over denominator up diagonal strike sin begin display style 3 over 2 end style up diagonal strike x times up diagonal strike sin 3 over 2 space up diagonal strike x times negative up diagonal strike tan 2 up diagonal strike x times tan 2 up diagonal strike x end fraction end cell row blank equals cell fraction numerator 3 times 3 times 3 times 3 over denominator begin display style fraction numerator 3 over denominator up diagonal strike 2 end fraction end style times begin display style fraction numerator 3 over denominator up diagonal strike 2 end fraction end style times up diagonal strike left parenthesis negative 2 right parenthesis end strike times up diagonal strike 2 end fraction end cell row blank equals cell fraction numerator up diagonal strike 3 times 3 end strike times 3 times 3 over denominator up diagonal strike 3 times 3 end strike times left parenthesis negative 1 right parenthesis end fraction end cell row blank equals cell fraction numerator 3 times 3 over denominator negative 1 end fraction end cell row blank equals cell negative 9 end cell end table$

Oleh karena itu, jawaban yang benar adalah E.