Nilai

Pertanyaan

Nilai $begin mathsize 14px style limit as x rightwards arrow infinity of fraction numerator open parentheses 1 minus 2 x close parentheses cubed over denominator open parentheses x minus 1 close parentheses open parentheses x squared plus x minus 1 close parentheses end fraction equals... end style$

$begin mathsize 14px style negative 8 end style$
$begin mathsize 14px style 8 end style$
$begin mathsize 14px style negative 2 end style$
$begin mathsize 14px style 2 end style$
$begin mathsize 14px style negative 1 end style$

Pembahasan Video:

Pembahasan Soal:

Perhatikan bahwa :

$\begin{array}{rcl}{{\color[rgb]{1.0, 0.0, 0.0}\mathbf(}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol a}{\color[rgb]{1.0, 0.0, 0.0}\mathbf-}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol b}{\color[rgb]{1.0, 0.0, 0.0}\mathbf)}}^{\color[rgb]{1.0, 0.0, 0.0}\mathbf3}&{\color[rgb]{1.0, 0.0, 0.0}\mathbf=}&{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol a}^{\color[rgb]{1.0, 0.0, 0.0}\mathbf3}{\color[rgb]{1.0, 0.0, 0.0}\mathbf-}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol b}^{\color[rgb]{1.0, 0.0, 0.0}\mathbf3}{\color[rgb]{1.0, 0.0, 0.0}\mathbf-}{\color[rgb]{1.0, 0.0, 0.0}\mathbf3}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol a}^{\color[rgb]{1.0, 0.0, 0.0}\mathbf2}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol b}{\color[rgb]{1.0, 0.0, 0.0}\mathbf+}{\color[rgb]{1.0, 0.0, 0.0}\mathbf3}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol a}{\color[rgb]{1.0, 0.0, 0.0}\boldsymbol b}^{\color[rgb]{1.0, 0.0, 0.0}\mathbf2}\\&&\\&&\mathrm{sehingga}\;:\\&&\\{(1-2x)}^3&=&1^3-{(2x)}^3-3\cdot1^2\cdot2x+3\cdot1\cdot{(2x)}^2\\&=&1-8x^3-6x+12x^2\end{array}$

Diperoleh :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator open parentheses 1 minus 2 x close parentheses cubed over denominator left parenthesis x minus 1 right parenthesis left parenthesis x squared plus x minus 1 right parenthesis end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator negative 8 x cubed plus 12 x squared minus 6 x plus 1 over denominator x cubed minus 2 x plus 1 end fraction end cell row blank blank blank end table$

Untuk menentukan limit di atas dapat menggunakan konsep limit tak hinggga bentuk fungsi rasional, yaitu dengan membagi pembilang dan penyebut dengan variabel yang memiliki pangkat tertinggi di penyebut

$Penyebut pada soal = x^{3} - 2 x + 1, maka variabel yang memiliki pangkat tertinggi adalah x^{3}$

Sehingga :

$table attributes columnalign right center left columnspacing 0px end attributes row cell space limit as x rightwards arrow infinity of fraction numerator negative 8 x cubed plus 12 x squared minus 6 x plus 1 over denominator x cubed minus 2 x plus 1 end fraction times fraction numerator begin display style 1 over x cubed end style over denominator begin display style 1 over x cubed end style end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator negative 8 plus begin display style 12 over x end style minus begin display style 6 over x squared end style plus begin display style 1 over x cubed end style over denominator 1 minus begin display style 2 over x squared end style plus begin display style 1 over x cubed end style end fraction space space space space space space space space bold lim with bold x bold rightwards arrow bold infinity below bold C over bold x to the power of bold n bold equals bold 0 bold space bold semicolon bold space bold italic n bold greater than bold 0 bold space bold space bold space end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end cell equals cell fraction numerator negative 8 plus 0 minus 0 plus 0 over denominator 1 minus 0 plus 0 end fraction end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end cell equals cell fraction numerator negative 8 over denominator 1 end fraction end cell row cell space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end cell equals cell negative 8 end cell end table$

Dengan demikian, Nilai $x \to \infty lim \frac{( 1 - 2 x ) ^{3}}{( x - 1 ) ( x ^{2} + x - 1 )}$ adalah $- 8$ .

Oleh karena itu, jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Hermawan

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 30 Agustus 2021

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Pembahasan Soal:

Jika diketahui fungsi $f open parentheses x close parentheses equals a x to the power of n plus a subscript 1 x to the power of n minus 1 end exponent plus...$ dengan pangkat tertinggi $n$ dan $g open parentheses x close parentheses equals b x to the power of m plus b subscript 1 x to the power of m minus 1 end exponent plus...$ dengan pangkat tertinggi $m$ , maka limit di tak hingganya:

$limit as x rightwards arrow infinity of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row 0 cell comma space n less than m end cell row cell a over b end cell cell comma space n equals m end cell row infinity cell comma space n greater than m end cell end table close$

Berdasarkan soal di atas, diketahui limit dengan pangkat tertinggi fungsi pembilangnya adalah $5$ dan pangkat tertinggi fungsi penyebutnya adalah $4$ . Artinya, pangkat tertinggi fungsi pembilang lebih besar dari pangkat tertingggi fungsi penyebut. Sehingga hasil limit adalah $infinity$ .

Jadi, $limit as x rightwards arrow infinity of fraction numerator x to the power of 5 minus 3 x squared minus 5 over denominator x to the power of 4 plus 2 x cubed plus 1 end fraction equals infinity$ .

Roboguru

Nilai dari adalah ....

Pembahasan Soal:

Untuk soal tersebut dapat memperhatikan ketentuan limit tak hingga berikut:

$limit as x rightwards arrow infinity of fraction numerator a x to the power of m plus b x to the power of m minus 1 end exponent plus c x to the power of m minus 2 end exponent plus horizontal ellipsis over denominator p x to the power of n plus q x to the power of n minus 1 end exponent plus r x to the power of n minus 2 end exponent plus horizontal ellipsis end fraction equals open curly brackets table attributes columnalign left end attributes row cell 0 comma space untuk space m less than n end cell row cell a over p comma space untuk space m equals n end cell row cell infinity comma space untuk space m greater than n end cell end table close$

Jika diperhatikan soal tersebut, diperoleh nilai m = 3 dan nilai n = 4. Artinya bahwa $m less than n$ sehingga berlaku syarat pertama.

$limit as x rightwards arrow infinity of fraction numerator 2 x cubed plus 3 x squared minus 5 x plus 4 over denominator 2 x to the power of 4 minus 4 x squared plus 9 end fraction equals 0$

Oleh karena itu, jawaban yang benar adalah B.

Roboguru

Pembahasan Soal:

Ingat kembali aturan limit tak hingga berikut ini.

Misal m dan n adalah bilangan bulat positif dan

$begin mathsize 12px style limit as x rightwards arrow infinity of fraction numerator a x to the power of m plus b x to the power of m minus 1 end exponent plus... plus c over denominator p x to the power of n plus q x to the power of n minus 1 end exponent plus... plus r end fraction equals L space rightwards arrow table attributes columnalign left end attributes row cell Jika space m greater than n space maka space L equals infinity space atau space minus infinity end cell row cell Jika space m equals n space maka space L equals a over p end cell row cell Jika space m less than n space maka space L equals 0 end cell end table end style$

Dengan aturan limit tak hingga di atas maka diperoleh perhitungan sebagai berikut.

Pada $limit as x rightwards arrow infinity of fraction numerator 6 x cubed plus x squared plus 2 over denominator 2 x cubed plus x plus 1 end fraction$ diperoleh bahwa

$m equals 3$
$n equals 3$
$a equals 6$
$p equals 2$

Karena $m italic equals n$ maka berlaku $L italic equals a over p$ .

Sehingga diperoleh

$limit as x rightwards arrow infinity of fraction numerator 6 x cubed plus x squared plus 2 over denominator 2 x cubed plus x plus 1 end fraction equals 6 over 2 equals 3$

Dengan demikian hasil $limit as x rightwards arrow infinity of fraction numerator 6 x cubed plus x squared plus 2 over denominator 2 x cubed plus x plus 1 end fraction equals 3$ .

Jadi, jawaban yang tepat adalah C.

Roboguru

Pembahasan Soal:

Perhatikan bentuk umum limit tak hingga berikut :

$limit as x rightwards arrow infinity of fraction numerator a x to the power of m plus b x to the power of m minus 1 end exponent plus horizontal ellipsis plus c over denominator p x to the power of n plus q x to the power of n minus 1 end exponent plus horizontal ellipsis plus r end fraction$ .

Penyelesaian limit di atas adalah :

jika $m greater than n$ maka $L equals infinity$ ,
jika $m equals n$ maka $L equals a over p$ , dan
jika $m less than n$ maka $L equals 0$ .

Pada soal diberikan :

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of space fraction numerator left parenthesis 2 x plus 1 right parenthesis left parenthesis 3 x plus 1 right parenthesis over denominator left parenthesis x plus 1 right parenthesis left parenthesis 4 x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator 6 x squared plus 2 x plus 3 x plus 1 over denominator 4 x squared minus x plus 4 x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator 6 x squared plus 5 x plus 1 over denominator 4 x squared plus 3 x minus 1 end fraction end cell end table$ .

Perhatikan bahwa pangkat tertinggi pada pembilang sama dengan pangkat tertinggi pada penyebut, atau $m equals n equals 2$ . Koefisien $a equals 6$ dan $p equals 4$ , sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of space fraction numerator 6 x squared plus 5 x plus 1 over denominator 4 x squared plus 3 x minus 1 end fraction end cell equals cell a over p equals 6 over 4 equals 1 1 half end cell end table$ .

Dengan demikian, hasil dari limit di atas adalah $1 1 half$ .

Jadi, jawaban yang tepat adalah D.

Roboguru

Pembahasan Soal:

Ingat salah satu rumus dasar limit berikut : $limit as x rightwards arrow infinity of space a over x to the power of n equals 0$ .

Diberikan bentuk limit : $limit as x rightwards arrow infinity of space fraction numerator square root of 2 x minus 1 end root plus square root of x plus 1 end root over denominator square root of 2 x plus 3 end root minus square root of x minus 2 end root end fraction$ .

Dengan mengalikan pembilang dan penyebut dengan $square root of 1 over x end root$ , maka bentuk di atas dapat diubah menjadi

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of space fraction numerator square root of 2 x minus 1 end root plus square root of x plus 1 end root over denominator square root of 2 x plus 3 end root minus square root of x minus 2 end root end fraction times fraction numerator square root of begin display style 1 over x end style end root over denominator square root of begin display style 1 over x end style end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator square root of begin display style 1 over x end style open parentheses 2 x minus 1 close parentheses end root plus square root of begin display style 1 over x end style open parentheses x plus 1 close parentheses end root over denominator square root of begin display style 1 over x end style left parenthesis 2 x plus 3 right parenthesis end root minus square root of begin display style 1 over x end style left parenthesis x minus 2 right parenthesis end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator square root of begin display style fraction numerator 2 x over denominator x end fraction end style minus begin display style 1 over x end style end root plus square root of begin display style x over x end style plus begin display style 1 over x end style end root over denominator square root of begin display style fraction numerator 2 x over denominator x end fraction end style plus begin display style 3 over x end style end root minus square root of begin display style x over x end style minus begin display style 2 over x end style end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator square root of 2 minus begin display style 1 over x end style end root plus square root of 1 plus begin display style 1 over x end style end root over denominator square root of 2 plus begin display style 3 over x end style end root minus square root of 1 minus begin display style 2 over x end style end root end fraction end cell row blank equals cell fraction numerator square root of 2 minus 0 end root plus square root of 1 plus 0 end root over denominator square root of 2 plus 0 end root minus square root of 1 minus 0 end root end fraction end cell row blank equals cell fraction numerator square root of 2 plus 1 over denominator square root of 2 minus 1 end fraction times fraction numerator square root of 2 plus 1 over denominator square root of 2 plus 1 end fraction end cell row blank equals cell fraction numerator open parentheses square root of 2 plus 1 close parentheses squared over denominator open parentheses square root of 2 close parentheses squared minus 1 squared end fraction end cell row blank equals cell fraction numerator 2 plus 2 square root of 2 plus 1 over denominator 2 minus 1 end fraction end cell row blank equals cell 3 plus 2 square root of 2 end cell end table$

Dengan demikian, hasil dari limit tersebut adalah $3 plus 2 square root of 2$ .

Jadi, jawaban yang tepat adalah A.

Roboguru