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Hasil dari x → 0 lim cos 7 x − cos 5 x 1 − cos 2 6 x adalah ....

Hasil dari $x \to 0 lim \frac{1 - cos ^{2} 6 x}{cos 7 x - cos 5 x}$ adalah ....

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V. Vierzaaa

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

Pembahasan

Berdasarkan identitas trigonometri, didapat sehingga . Selanjutnya, dapat diperhatikan bahwa . Dengan demikian, didapat perhitungan sebagai berikut. Jadi, jawaban yang tepat adalah A.

Berdasarkan identitas trigonometri, didapat sehingga .

Selanjutnya, dapat diperhatikan bahwa $size 14px cos size 14px invisible function application size 14px alpha size 14px minus size 14px cos size 14px invisible function application size 14px beta size 14px equals size 14px minus size 14px 2 size 14px sin size 14px invisible function application open parentheses fraction numerator size 14px alpha size 14px plus size 14px beta over denominator size 14px 2 end fraction close parentheses size 14px sin size 14px invisible function application open parentheses fraction numerator size 14px alpha size 14px minus size 14px beta over denominator size 14px 2 end fraction close parentheses$ .

Dengan demikian, didapat perhitungan sebagai berikut.

$begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application fraction numerator 1 minus cos squared invisible function application 6 x over denominator cos invisible function application 7 x minus cos invisible function application 5 x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 1 minus open parentheses 1 minus sin squared invisible function application 6 x close parentheses over denominator negative 2 sin invisible function application open parentheses fraction numerator 7 x plus 5 x over denominator 2 end fraction close parentheses sin invisible function application open parentheses fraction numerator 7 x minus 5 x over denominator 2 end fraction close parentheses end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin squared invisible function application 6 x over denominator negative 2 sin invisible function application 6 x sin invisible function application x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 6 x sin invisible function application 6 x over denominator negative 2 sin invisible function application 6 x sin invisible function application x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 6 x over denominator negative 2 sin invisible function application x end fraction equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator sin invisible function application 6 x over denominator negative 2 sin invisible function application x end fraction times fraction numerator 6 x over denominator 6 x end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator sin invisible function application 6 x over denominator 6 x end fraction times fraction numerator x over denominator sin invisible function application x end fraction times fraction numerator 6 over denominator negative 2 end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application fraction numerator sin invisible function application 6 x over denominator 6 x end fraction times limit as x rightwards arrow 0 of invisible function application fraction numerator x over denominator sin invisible function application x end fraction times limit as x rightwards arrow 0 of invisible function application fraction numerator 6 over denominator negative 2 end fraction equals 1 times 1 times open parentheses negative 3 close parentheses equals negative 3 end style$

Jadi, jawaban yang tepat adalah A.

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