Pertanyaan

Nilai dari adalah....

Nilai dari $begin mathsize 14px style lim with x yields infinity below invisible function application fraction numerator open parentheses 2 sin squared invisible function application open parentheses 2 over x close parentheses close parentheses open parentheses sec squared invisible function application open parentheses 2 over x close parentheses close parentheses over denominator x to the power of negative sign 2 end exponent end fraction end style$ adalah....

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Jawaban terverifikasi

Pembahasan

Perhatikan bahwa Sehingga Misalkan ,maka .Ketika , maka . Sehingga Maka, jawaban yang tepat adalah D.

Perhatikan bahwa

$begin mathsize 14px style sec squared invisible function application open parentheses 2 over x close parentheses equals fraction numerator 1 over denominator cos squared invisible function application open parentheses 2 over x close parentheses end fraction end style$

Sehingga

Misalkan , maka . Ketika , maka .

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight infinity of invisible function application open parentheses 2 x squared tan squared invisible function application 2 open parentheses 1 over x close parentheses close parentheses end cell equals cell limit as y rightwards arrow 0 of invisible function application open parentheses 2 times open parentheses 1 over y close parentheses squared tan squared invisible function application 2 y close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 tan squared invisible function application 2 y over denominator y squared end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 tan invisible function application 2 y times tan invisible function application 2 y over denominator y times y end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 tan invisible function application 2 y over denominator y end fraction times fraction numerator tan invisible function application 2 y over denominator y end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses fraction numerator 2 tan invisible function application 2 y over denominator y end fraction times 2 over 2 times fraction numerator tan invisible function application 2 y over denominator y end fraction times 2 over 2 close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of invisible function application open parentheses 2 times fraction numerator tan invisible function application 2 y over denominator 2 y end fraction times 2 times fraction numerator tan invisible function application 2 y over denominator 2 y end fraction times 2 close parentheses end cell row blank equals cell 2 times 1 times 2 times 1 times 2 end cell row blank equals cell space 8 end cell end table end style$

Maka, jawaban yang tepat adalah D.