Tentukan nilai x→∞lim3x+2 sin x−8cos 4x− 6x!

Pertanyaan

Tentukan nilai $limit as x rightwards arrow infinity of fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction$ !

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

nilai dari $limit as x rightwards arrow infinity of fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction$ adalah $-2$ .

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah $\boldsymbol-\mathbf2$ .

Ingat!

nilai $\cos\;x$ , selalu berada di antara $-1$ dan $1$
nilai $\sin\;x$ , selalu berada di antara $-1$ dan $1$
$\lim_{x\rightarrow\infty}\frac k{x^n}=0,\;\text{dimana}\;n\text{ bilangan asli}$
sifat limit: $\lim_{x\rightarrow0}(f(x):f(g))=\lim_{x\rightarrow0}f(x):\lim_{x\rightarrow0}f(g)$

Apabila nilai $x=\infty$ langsung di subsitusi ke fungsi $fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction$ maka akan di peroleh nilai limit $\frac\infty\infty$ . Oleh karena itu harus disederhanakan terlebih dahulu.

Kalikan pembilang dan penyebut dengan $\frac1x$ , sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction times fraction numerator begin display style bevelled 1 over x end style over denominator begin display style bevelled 1 over x end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style fraction numerator cos space 4 x over denominator x end fraction end style minus space begin display style fraction numerator 6 x over denominator x end fraction end style over denominator begin display style fraction numerator 3 x over denominator x end fraction end style plus begin display style fraction numerator 2 space sin space x over denominator x end fraction end style minus begin display style 8 over x end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style fraction numerator cos space 4 x over denominator x end fraction minus space 6 end style over denominator begin display style 3 plus fraction numerator 2 space sin space x over denominator x end fraction minus 8 over x end style end fraction end cell end table$

Nilai $\cos\;x$ , selalu berada di antara $-1$ dan $1$ , maka $-1\leq\cos\;4x\leq1$ , sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative 1 end cell less or equal than cell cos space 4 x less or equal than 1 end cell row cell fraction numerator negative 1 over denominator x end fraction end cell less or equal than cell fraction numerator cos space 4 x over denominator x end fraction less or equal than 1 over x end cell row cell limit as x rightwards arrow infinity of fraction numerator negative 1 over denominator x end fraction end cell less or equal than cell limit as x rightwards arrow infinity of fraction numerator cos space 4 x over denominator x end fraction less or equal than limit as x rightwards arrow infinity of 1 over x end cell row 0 less or equal than cell limit as x rightwards arrow infinity of fraction numerator cos space 4 x over denominator x end fraction less or equal than 0 end cell end table$

Oleh karena itu $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator cos space 4 x over denominator x end fraction end cell end table equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table$ .

Begitu juga dengan nilai $\sin\;x$ , selalu berada di antara $-1$ dan $1$ , maka $-1\leq\sin\;x\leq1$ . Sehingga:

$table attributes columnalign right center left columnspacing 0px end attributes row cell negative 1 end cell less or equal than cell sin space x less or equal than 1 end cell row cell fraction numerator negative 1 over denominator x end fraction end cell less or equal than cell fraction numerator sin space x over denominator x end fraction less or equal than 1 over x end cell row cell limit as x rightwards arrow infinity of fraction numerator negative 1 over denominator x end fraction end cell less or equal than cell limit as x rightwards arrow infinity of fraction numerator sin space x over denominator x end fraction less or equal than limit as x rightwards arrow infinity of 1 over x end cell row 0 less or equal than cell limit as x rightwards arrow infinity of fraction numerator sin space x over denominator x end fraction less or equal than 0 end cell row blank blank blank end table$

Oleh karena itu $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator sin space x over denominator x end fraction end cell end table equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table$ . Sehingga di peroleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator begin display style fraction numerator cos space 4 x over denominator x end fraction minus space 6 end style over denominator begin display style 3 plus fraction numerator 2 space sin space x over denominator x end fraction minus 8 over x end style end fraction end cell equals cell fraction numerator begin display style limit as x rightwards arrow infinity of fraction numerator cos space 4 x over denominator x end fraction minus limit as x rightwards arrow infinity of space 6 end style over denominator begin display style limit as x rightwards arrow infinity of 3 plus 2 limit as x rightwards arrow infinity of fraction numerator sin space x over denominator x end fraction minus limit as x rightwards arrow infinity of 8 over x end style end fraction end cell row blank equals cell fraction numerator 0 minus 6 over denominator 3 plus 0 plus 0 end fraction end cell row blank equals cell fraction numerator negative 6 over denominator 3 end fraction end cell row blank equals cell negative 2 end cell end table$

Dengan demikian nilai dari $limit as x rightwards arrow infinity of fraction numerator cos space 4 x minus space 6 x over denominator 3 x plus 2 space sin space x minus 8 end fraction$ adalah $-2$ .