Pertanyaan

Hasil dari x → 0 lim 2 cos 4 x − 2 5 x tan 2 x adalah ...

Hasil dari $x \to 0 lim \frac{5 x tan 2 x}{2 cos 4 x - 2}$ adalah ...

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Habis dalam

Klaim

N. Mustikowati

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan bahwa Jika disubstitusi , maka didapat Oleh karena itu, didapat Kemudian, didapat bahwa Jadi, jawaban yang tepat adalah E.

Perhatikan bahwa

Jika disubstitusi , maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application 4 x end cell equals cell 1 minus 2 sin squared invisible function application open parentheses 1 half times 4 x close parentheses end cell row blank equals cell 1 minus 2 sin squared invisible function application 2 x end cell end table end style

Oleh karena itu, didapat

Kemudian, didapat bahwa

$begin mathsize 14px style limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator 2 cos invisible function application 4 x minus 2 end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator negative 4 sin squared invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x tan invisible function application 2 x over denominator negative 4 sin invisible function application 2 x sin invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application fraction numerator 5 x over denominator negative 4 sin invisible function application 2 x end fraction times limit as x rightwards arrow 0 of invisible function application fraction numerator tan invisible function application 2 x over denominator sin invisible function application 2 x end fraction equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator 5 x over denominator negative 4 sin invisible function application 2 x end fraction times 2 over 2 close parentheses times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 2 x over denominator sin invisible function application 2 x end fraction times fraction numerator 2 x over denominator 2 x end fraction close parentheses equals limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 x over denominator sin invisible function application 2 x end fraction times fraction numerator 5 over denominator negative 4 times 2 end fraction close parentheses times limit as x rightwards arrow 0 of invisible function application open parentheses fraction numerator tan invisible function application 2 x over denominator 2 x end fraction times fraction numerator 2 x over denominator sin invisible function application 2 x end fraction close parentheses equals open parentheses 1 times fraction numerator 5 over denominator negative 8 end fraction close parentheses times open parentheses 1 times 1 close parentheses equals negative 5 over 8 end style$

Jadi, jawaban yang tepat adalah E.