Pertanyaan

x → ∞ lim 8 x 2 − 1 ( 2 x + 3 ) 2 − 7 = ...

$x \to \infty lim \frac{( 2 x + 3 ) ^{2} - 7}{8 x ^{2} - 1} = ...$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Ingat: dan Jadi Oleh karena itu, jawaban yang benar adalah C.

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.