Tentukan nilai dari: f.

Pertanyaan

Tentukan nilai dari:

f. $begin mathsize 14px style limit as x rightwards arrow infinity of fraction numerator 8 x squared plus 2 x minus 1 over denominator 2 x plus 4 end fraction end style$

Pembahasan Video:

Pembahasan Soal:

Dengan menggunakan konsep limit tak hingga

$Error converting from MathML to accessible text.$

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 8 x squared plus 2 x minus 1 over denominator 2 x plus 4 end fraction end cell end table$ mempunyai pangkat tertinggi pada pembilangnya 2 dan pangkat tertinggi pada penyebutnya 1, sehingga berdasarkan rumus diatas $m greater than n$ maka :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator 8 x squared plus 2 x minus 1 over denominator 2 x plus 4 end fraction end cell equals infinity end table$

Jadi, hasil limitnya adalah $\infty$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 30 Agustus 2021

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Pembahasan Soal:

Ingat: $limit as x rightwards arrow infinity of k over x to the power of n equals k over infinity to the power of n equals 0$ dan $open parentheses a plus b close parentheses squared equals a squared plus 2 a b plus b squared$

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator open parentheses 2 x plus 3 close parentheses squared minus 7 over denominator 8 x squared minus 1 end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator open parentheses 2 x close parentheses squared plus 2 times 2 x times 3 plus 3 squared minus 7 over denominator 8 x squared minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 4 x squared plus 12 x plus 2 over denominator 8 x squared minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator 4 x squared plus 12 x plus 2 over denominator 8 x squared minus 1 end fraction cross times fraction numerator begin display style 1 over x squared end style over denominator begin display style 1 over x squared end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style fraction numerator 4 x squared over denominator x squared end fraction end style plus begin display style fraction numerator 12 x over denominator x squared end fraction end style plus begin display style 2 over x squared end style over denominator begin display style fraction numerator 8 x squared over denominator x squared end fraction end style minus begin display style 1 over x squared end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style 4 plus 12 over x plus 2 over x squared end style over denominator begin display style 8 minus 1 over x squared end style end fraction end cell row blank equals cell fraction numerator 4 plus begin display style 12 over infinity end style plus begin display style 2 over infinity squared end style over denominator 8 minus begin display style 1 over infinity squared end style end fraction end cell row blank equals cell fraction numerator 4 plus 0 plus 0 over denominator 8 minus 0 end fraction end cell row blank equals cell 0 comma 5 end cell end table$

Jadi $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow infinity of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator open parentheses 2 x plus 3 close parentheses squared minus 7 over denominator 8 x squared minus 1 end fraction end cell end table equals 0 comma 5$

Oleh karena itu, jawaban yang benar adalah C.

Roboguru

Pembahasan Soal:

Ingat: $limit as x rightwards arrow infinity of k over x to the power of n equals k over infinity to the power of n equals 0$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator x squared left parenthesis x squared minus 2 x plus 1 right parenthesis over denominator x squared plus 4 x plus 5 end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator x to the power of 4 minus 2 x cubed plus x squared over denominator x squared plus 4 x plus 5 end fraction times fraction numerator begin display style 1 over x to the power of 4 end style over denominator begin display style 1 over x to the power of 4 end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style x to the power of 4 over x to the power of 4 end style minus begin display style fraction numerator 2 x cubed over denominator x to the power of 4 end fraction end style plus begin display style x squared over x to the power of 4 end style over denominator begin display style x squared over x to the power of 4 end style plus begin display style fraction numerator 4 x over denominator x to the power of 4 end fraction end style plus begin display style 5 over x to the power of 4 end style end fraction end cell row blank equals cell limit as x rightwards arrow infinity of fraction numerator begin display style 1 minus 2 over x plus 1 over x squared end style over denominator begin display style 1 over x squared plus 4 over x cubed plus 5 over x to the power of 4 end style end fraction end cell row blank equals cell fraction numerator begin display style 1 minus 2 over infinity plus 1 over infinity squared end style over denominator begin display style 1 over infinity squared plus 4 over infinity cubed plus 5 over infinity to the power of 4 end style end fraction end cell row blank equals cell fraction numerator 1 minus 0 plus 0 over denominator 0 plus 0 plus 0 end fraction end cell row blank equals cell 1 over 0 end cell row blank equals infinity end table$

Dengan demikian, nilai dari limit di atas adalah $begin mathsize 14px style infinity end style$ .

Jadi, jawaban yang tepat adalah E.

Roboguru

Nilai dari adalah...

Pembahasan Soal:

Perhatikan penghitungan berikut!

$limit as x rightwards arrow infinity of space fraction numerator square root of 5 x plus 2020 end root minus square root of 4 x plus 2021 end root over denominator 4 x end fraction equals limit as x rightwards arrow infinity of space fraction numerator square root of begin display style fraction numerator 5 x over denominator x squared end fraction end style plus begin display style 2020 over x squared end style end root minus square root of begin display style fraction numerator 4 x over denominator x squared end fraction end style plus begin display style 2021 over x squared end style end root over denominator begin display style fraction numerator 4 x over denominator x end fraction end style end fraction equals limit as x rightwards arrow infinity of space fraction numerator square root of begin display style 5 over x plus 2020 over x squared end style end root minus square root of begin display style 4 over x plus 2021 over x squared end style end root over denominator begin display style 4 end style end fraction equals fraction numerator square root of begin display style 0 plus 0 end style end root minus square root of begin display style 0 plus 0 end style end root over denominator begin display style 4 end style end fraction equals 0 over 4 equals 0$

Jadi, nilai dari $limit as x rightwards arrow infinity of space fraction numerator square root of 5 x plus 2020 end root minus square root of 4 x plus 2021 end root over denominator 4 x end fraction$ adalah 0. $space$

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Pembahasan Soal:

Ingat kembali :

$Error converting from MathML to accessible text.$

$limit as x rightwards arrow infinity of fraction numerator 4 x to the power of 5 plus 3 x squared minus 5 x plus 7 over denominator 2 x to the power of 5 minus 8 x cubed plus 3 x minus 2 end fraction$ mempunyai pangkat tertinggi dari penyebut dan pembilangnya sama yaitu 5, sehingga berdasarkan rumus diatas :

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator 4 x to the power of 5 plus 3 x squared minus 5 x plus 7 over denominator 2 x to the power of 5 minus 8 x cubed plus 3 x minus 2 end fraction end cell equals cell 4 over 2 end cell row blank equals 2 end table$

Jadi, nilai dari $limit as x rightwards arrow infinity of fraction numerator 4 x to the power of 5 plus 3 x squared minus 5 x plus 7 over denominator 2 x to the power of 5 minus 8 x cubed plus 3 x minus 2 end fraction$ adalah $begin mathsize 14px style 2 end style$ .

Roboguru

Pembahasan Soal:

Ingat: $limit as x rightwards arrow infinity of k over x to the power of n equals k over infinity to the power of n equals 0$ .

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

Jadi, hasil limitnya adalah 0.

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