Pertanyaan

x → 0 lim x 3 sin 2 x + 1 − tan 2 x + 1 = ...

$x \to 0 lim \frac{sin 2 x + 1 - tan 2 x + 1}{x ^{3}} = ...$

-8
-4
-2
2
4

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I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

$begin mathsize 12px style limit as x rightwards arrow 0 of fraction numerator square root of sin space 2 x plus 1 end root minus square root of tan space 2 x plus 1 end root over denominator x cubed end fraction. open parentheses fraction numerator square root of sin space 2 x plus 1 end root plus square root of tan space 2 x plus 1 end root over denominator square root of sin space 2 x plus 1 end root plus square root of tan space 2 x plus 1 end root end fraction close parentheses limit as x rightwards arrow 0 of fraction numerator open parentheses sin space 2 x plus 1 close parentheses minus open parentheses tan space 2 x plus 1 close parentheses over denominator x cubed open parentheses square root of sin space 2 x plus 1 end root plus square root of tan space 2 x plus 1 end root close parentheses end fraction limit as x rightwards arrow 0 of fraction numerator sin space 2 x minus tan space 2 x over denominator x cubed open parentheses square root of 1 plus square root of 1 close parentheses end fraction limit as x rightwards arrow 0 of fraction numerator sin space 2 x minus tan space 2 x over denominator 2 x cubed end fraction end style$

$begin mathsize 12px style Ingat space bahwa space tan space 2 straight x equals fraction numerator sin space 2 straight x over denominator cos space 2 straight x end fraction left right double arrow sin space 2 straight x equals tan space 2 straight x. cos space 2 straight x limit as straight x rightwards arrow 0 of fraction numerator tan space 2 straight x. cos space 2 straight x minus tan space 2 straight x over denominator 2 straight x cubed end fraction limit as straight x rightwards arrow 0 of fraction numerator tan space 2 straight x. open parentheses cos space 2 straight x minus 1 close parentheses over denominator 2 straight x cubed end fraction Ingat space cos space 2 straight x equals 1 minus 2 sin squared space straight x limit as straight x rightwards arrow 0 of fraction numerator tan space 2 straight x. open parentheses negative 2 sin squared space straight x close parentheses over denominator 2 straight x cubed end fraction limit as straight x rightwards arrow 0 of fraction numerator negative 2. tan space 2 straight x. sin space straight x. sin space straight x over denominator 2 straight x cubed end fraction equals negative 2 end style$

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