x→∞lim x2+2x−13x−1=…

Pertanyaan

$limit as x rightwards arrow infinity of space fraction numerator 3 x minus 1 over denominator square root of x squared plus 2 x minus 1 end root end fraction equals horizontal ellipsis$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

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

Dengan mengalikan pembilang dan penyebut dengan 1 over x , maka bentuk di atas dapat diubah menjadi

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

Dengan demikian, hasil dari limit tersebut adalah .

Jadi, jawaban yang tepat adalah D.