Pertanyaan

Nilai dari adalah ...

Nilai dari $begin mathsize 14px style limit as straight x rightwards arrow infinity of fraction numerator 2 straight x space cot space open parentheses begin display style 2 over straight x end style close parentheses minus 3 space cot space open parentheses begin display style 2 over straight x end style close parentheses over denominator 5 straight x squared minus 2 straight x end fraction end style$ adalah ...

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Misalkan , sehingga . Ketika , maka . Maka Jadi, jawaban yang tepat adalah C.

Misalkan begin mathsize 14px style 1 over straight x equals straight y end style , sehingga $begin mathsize 14px style cot space straight y equals fraction numerator 1 over denominator tan space straight y end fraction end style$ . Ketika begin mathsize 14px style straight x rightwards arrow infinity end style , maka .

Maka

$begin mathsize 14px style equals limit as straight x rightwards arrow infinity of fraction numerator 2 straight x space cot space open parentheses begin display style 2 over straight x end style close parentheses minus 3 space cot space open parentheses begin display style 2 over straight x end style close parentheses over denominator 5 straight x squared minus 2 straight x end fraction equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses space cot space open parentheses begin display style 2 over straight x end style close parentheses over denominator straight x space open parentheses 5 straight x minus 2 close parentheses end fraction equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses over denominator open parentheses 5 straight x minus 2 close parentheses end fraction space. space 1 over straight x space. space cot space open parentheses 2 over straight x close parentheses equals limit as straight x rightwards arrow infinity of fraction numerator open parentheses 2 straight x minus 3 close parentheses over denominator open parentheses 5 straight x minus 2 close parentheses end fraction space. space limit as straight x rightwards arrow infinity of 1 over straight x space. space cot space open parentheses 2 over straight x close parentheses equals 2 over 5 space. space limit as straight y rightwards arrow 0 of straight y space. space cot space 2 straight y equals 2 over 5 space. space limit as straight y rightwards arrow 0 of straight y space. space fraction numerator 1 over denominator tan space 2 straight y end fraction equals 2 over 5 space. space limit as straight y rightwards arrow 0 of space fraction numerator straight y over denominator tan space 2 straight y end fraction equals 2 over 5 space. space 1 half equals 1 fifth end style$

Jadi, jawaban yang tepat adalah C.