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Hitung dan tentukan limit berikut ini! x → ∞ lim ​ x 3 sin x 1 ​ ( sec x 1 ​ − 1 ) = ...

Hitung dan tentukan limit berikut ini! 

  1. 2

  2. 1

  3. 1 half

  4. 1 fourth

  5. 1 over 8

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H. Eka

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

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Pembahasan

Misalkan

Misalkan 1 over straight x equals straight y comma space maka space straight x equals 1 over straight y 

Untuk space straight x rightwards arrow infinity comma space maka space straight y rightwards arrow 0 limit as straight x rightwards arrow infinity of straight x cubed sin 1 over straight x left parenthesis sec 1 over straight x minus 1 right parenthesis limit as straight y rightwards arrow 0 of open parentheses 1 over straight y close parentheses cubed sin space straight y left parenthesis sec space straight y minus 1 right parenthesis limit as straight y rightwards arrow 0 of open parentheses 1 over straight y close parentheses cubed sin space straight y left parenthesis fraction numerator 1 over denominator cos space straight y end fraction minus 1 right parenthesis limit as straight y rightwards arrow 0 of open parentheses 1 over straight y close parentheses cubed sin space straight y left parenthesis fraction numerator 1 minus cos space straight y over denominator cos space straight y end fraction right parenthesis limit as straight y rightwards arrow 0 of 1 over straight y cubed sin space straight y open parentheses fraction numerator 1 minus left parenthesis 1 minus 2 sin squared begin display style 1 half end style straight y right parenthesis over denominator cos space straight y end fraction close parentheses limit as straight y rightwards arrow 0 of 1 over straight y cubed sin space straight y open parentheses fraction numerator 2 sin squared begin display style 1 half end style straight y right parenthesis over denominator cos space straight y end fraction close parentheses limit as straight y rightwards arrow 0 of fraction numerator 2 sin squared begin display style 1 half end style straight y. space sin space straight y over denominator straight y cubed. space cos space straight y end fraction limit as straight y rightwards arrow 0 of fraction numerator 2 begin display style 1 half end style begin display style 1 half end style over denominator space cos space straight y end fraction equals fraction numerator begin display style 1 half end style over denominator 1 end fraction equals 1 half

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x → ∞ lim ​ x ( sec x ​ 1 ​ − 1 ) = .... (SBMPTN 207)

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