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x→∞lim​ 4x2+x+4​+x2+1​3x−1​=…

Pertanyaan

limit as x rightwards arrow infinity of space fraction numerator 3 x minus 1 over denominator square root of 4 x squared plus x plus 4 end root plus square root of x squared plus 1 end root end fraction equals horizontal ellipsis  

  1. 0 

  2. 3 over 5 

  3. 1 third square root of 3 

  4. 1 half 

  5. 1 

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

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 3 x minus 1 over denominator square root of 4 x squared plus x plus 4 end root plus square root of x squared plus 1 end root end fraction.

Dengan mengalikan pembilang dan penyebut dengan 1 over x, 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 3 x minus 1 over denominator square root of 4 x squared plus x plus 4 end root plus square root of x squared plus 1 end root end fraction 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 begin display style fraction numerator 3 x over denominator x end fraction end style minus begin display style 1 over x end style over denominator begin display style 1 over x end style open parentheses square root of 4 x squared plus x plus 4 end root plus square root of x squared plus 1 end root close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator 3 minus begin display style 1 over x end style over denominator square root of begin display style 1 over x squared end style open parentheses 4 x squared plus x plus 4 close parentheses end root plus square root of begin display style 1 over x squared end style open parentheses x squared plus 1 close parentheses end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator 3 minus begin display style 1 over x end style over denominator square root of begin display style fraction numerator 4 x squared over denominator x squared end fraction end style plus begin display style x over x squared end style plus begin display style 4 over x squared end style end root plus square root of begin display style x squared over x squared end style plus begin display style 1 over x squared end style end root end fraction end cell row blank equals cell limit as x rightwards arrow infinity of space fraction numerator 3 minus begin display style 1 over x end style over denominator square root of 4 plus begin display style 1 over x end style plus begin display style 4 over x squared end style end root plus square root of 1 plus begin display style 1 over x squared end style end root end fraction end cell row blank equals cell fraction numerator 3 minus 0 over denominator square root of 4 plus 0 plus 0 end root plus square root of 1 plus 0 end root end fraction end cell row blank equals cell fraction numerator 3 over denominator square root of 4 plus square root of 1 end fraction end cell row blank equals cell fraction numerator 3 over denominator 2 plus 1 end fraction end cell row blank equals cell 3 over 3 end cell row blank equals 1 end table 

Dengan demikian, hasil dari limit tersebut adalah 1.

Jadi, jawaban yang tepat adalah E.

164

5.0 (8 rating)

Maria Aurelia Permata Sari

Pembahasan lengkap banget Makasih ❤️

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