Pertanyaan

x → 0 lim tan 2 x + sin x x + sin 2 x + sin 3 x = ...

$x \to 0 lim \frac{x + sin 2 x + sin 3 x}{tan 2 x + sin x} = ...$

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Jawaban terverifikasi

Pembahasan

Ingat kembalisifat limit dan limit fungsi trigonometri berikut. Dari aturan di atas, maka diperoleh Dengan demikian, . Jadi, jawaban yang benar adalah A.

Ingat kembali sifat limit dan limit fungsi trigonometri berikut.

$limit as x rightwards arrow c of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator limit as x rightwards arrow c of space f left parenthesis x right parenthesis over denominator limit as x rightwards arrow c of space g left parenthesis x right parenthesis end fraction$
$limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals a over b$
$limit as x rightwards arrow 0 of fraction numerator tan space a x over denominator b x end fraction equals a over b$

Dari aturan di atas, maka diperoleh

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator x plus sin space 2 x plus sin space 3 x over denominator tan space 2 x plus sin space x end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator open parentheses x plus sin space 2 x plus sin space 3 x close parentheses over denominator open parentheses tan space 2 x plus sin space x close parentheses 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 0 of fraction numerator begin display style x over x end style plus begin display style fraction numerator sin space 2 x over denominator x end fraction end style plus begin display style fraction numerator sin space 3 x over denominator x end fraction end style over denominator begin display style fraction numerator tan space 2 x over denominator x end fraction end style plus begin display style fraction numerator sin space x over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 0 of open parentheses space x over x plus fraction numerator sin space 2 x over denominator x end fraction plus fraction numerator sin space 3 x over denominator x end fraction close parentheses end style over denominator begin display style limit as x rightwards arrow 0 of space open parentheses fraction numerator tan space 2 x over denominator x end fraction plus fraction numerator sin space x over denominator x end fraction close parentheses end style end fraction end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 0 of space x over x plus limit as x rightwards arrow 0 of fraction numerator sin space 2 x over denominator x end fraction plus limit as x rightwards arrow 0 of fraction numerator sin space 3 x over denominator x end fraction end style over denominator begin display style limit as x rightwards arrow 0 of space fraction numerator tan space 2 x over denominator x end fraction plus limit as x rightwards arrow 0 of fraction numerator sin space x over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator 1 plus 2 plus 3 over denominator 2 plus 1 end fraction end cell row blank equals cell 6 over 3 end cell row blank equals 2 end table end style$

Dengan demikian, $limit as x rightwards arrow 0 of fraction numerator x plus sin space 2 x plus sin space 3 x over denominator tan space 2 x plus sin space x end fraction equals 2$ .

Jadi, jawaban yang benar adalah A.