Pertanyaan

Nilai adalah ....

Nilai $stack text lim end text with x rightwards arrow negative 3 below space fraction numerator x squared plus 6 x plus 9 over denominator 2 minus 2 space cos space open parentheses 2 x plus 6 close parentheses end fraction$ adalah ....

$3$
$1$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{4}$

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Habis dalam

Klaim

A. Azriah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah E.

Pembahasan

Ingat persamaan trigonometri dan sifat limit trigonometri berikut! 1 − cos x = 2 sin 2 2 1 x x → 0 lim sin x x = 1 Penyelesaian: Oleh karena itu, jawaban yang benar adalah E.

Ingat persamaan trigonometri dan sifat limit trigonometri berikut!

$1 - cos x = 2 sin^{2} \frac{1}{2} x$
$x \to 0 lim \frac{x}{sin x} = 1$

Penyelesaian:

$table attributes columnalign right center left columnspacing 0px end attributes row cell stack text lim end text with x rightwards arrow negative 3 below fraction numerator x squared plus 6 x plus 9 over denominator 2 minus 2 space cos space open parentheses 2 x plus 6 close parentheses end fraction end cell equals cell stack text lim end text with x rightwards arrow negative 3 below fraction numerator open parentheses x plus 3 close parentheses squared over denominator 2 open parentheses 1 minus cos space left parenthesis 2 x plus 6 right parenthesis close parentheses end fraction end cell row blank equals cell 1 half stack text lim end text with x rightwards arrow negative 3 below fraction numerator open parentheses x plus 3 close parentheses squared over denominator open parentheses 1 minus cos space left parenthesis 2 x plus 6 right parenthesis close parentheses end fraction end cell row blank equals cell 1 half stack text lim end text with x rightwards arrow negative 3 below fraction numerator open parentheses x plus 3 close parentheses squared over denominator 2 space sin squared space open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell 1 half times 1 half stack text lim end text with x rightwards arrow negative 3 below fraction numerator open parentheses x plus 3 close parentheses squared over denominator sin squared space open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell 1 fourth stack text lim end text with x rightwards arrow negative 3 below open parentheses fraction numerator left parenthesis x plus 3 right parenthesis over denominator sin space open parentheses x plus 3 close parentheses end fraction close parentheses squared end cell row blank equals cell 1 fourth open parentheses 1 close parentheses squared end cell row blank equals cell 1 fourth end cell end table$

Oleh karena itu, jawaban yang benar adalah E.