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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Misalkan maka untuk maka . Ingat pula bahwa . Kemudian, perhatikan perhitungan berikut! Jadi, jawaban yang tepat adalah B.

$limit as x rightwards arrow infinity of invisible function application x open parentheses sec invisible function application fraction numerator 1 over denominator square root of x end fraction minus 1 close parentheses limit as x rightwards arrow infinity of invisible function application x open parentheses fraction numerator 1 over denominator cos invisible function application open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses end fraction minus 1 close parentheses$

$limit as x rightwards arrow infinity of invisible function application x open parentheses fraction numerator 1 minus cos invisible function application open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses over denominator cos invisible function application open parentheses fraction numerator 1 over denominator square root of x end fraction close parentheses end fraction close parentheses$

Misalkan $u equals fraction numerator 1 over denominator square root of x end fraction$ maka untuk maka .

Ingat pula bahwa $begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator sin x over denominator x end fraction equals 1 end style$ . Kemudian, perhatikan perhitungan berikut!

$begin mathsize 14px style limit as u rightwards arrow 0 of invisible function application 1 over u squared open parentheses fraction numerator 1 minus cos invisible function application u over denominator cos invisible function application u end fraction close parentheses equals limit as u rightwards arrow 0 of invisible function application 1 over u squared open parentheses fraction numerator 1 minus cos invisible function application u over denominator cos invisible function application u end fraction close parentheses. fraction numerator 1 plus cos invisible function application u over denominator 1 plus cos invisible function application u end fraction equals limit as u rightwards arrow 0 of invisible function application fraction numerator 1 minus cos squared invisible function application u over denominator u squared open parentheses cos invisible function application u close parentheses left parenthesis 1 plus cos invisible function application u right parenthesis end fraction equals limit as u rightwards arrow 0 of invisible function application fraction numerator sin squared invisible function application u over denominator u squared open parentheses cos invisible function application u close parentheses left parenthesis 1 plus cos invisible function application u right parenthesis end fraction equals limit as u rightwards arrow 0 of fraction numerator sin u times sin u over denominator u times u times open parentheses cos invisible function application u close parentheses left parenthesis 1 plus cos invisible function application u right parenthesis end fraction equals limit as u rightwards arrow 0 of fraction numerator sin u over denominator u end fraction times fraction numerator sin u over denominator u end fraction times fraction numerator 1 over denominator open parentheses cos invisible function application u close parentheses left parenthesis 1 plus cos invisible function application u right parenthesis end fraction equals limit as u rightwards arrow 0 of fraction numerator sin u over denominator u end fraction times limit as u rightwards arrow 0 of fraction numerator sin u over denominator u end fraction times limit as u rightwards arrow 0 of fraction numerator 1 over denominator open parentheses cos invisible function application u close parentheses left parenthesis 1 plus cos invisible function application u right parenthesis end fraction equals 1 times 1 times fraction numerator 1 over denominator open parentheses cos invisible function application 0 close parentheses left parenthesis 1 plus cos invisible function application 0 right parenthesis end fraction equals fraction numerator 1 over denominator open parentheses cos invisible function application 0 close parentheses left parenthesis 1 plus cos invisible function application 0 right parenthesis end fraction equals fraction numerator 1 over denominator open parentheses 1 close parentheses open parentheses 1 plus 1 close parentheses end fraction equals fraction numerator 1 over denominator 1 times 2 end fraction equals 1 half end style$

Jadi, jawaban yang tepat adalah B.

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