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Hasil dari adalah ....

Hasil dari $begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses x squared plus 4 x minus 5 close parentheses over denominator sin invisible function application open parentheses x squared minus 1 close parentheses end fraction end style$ adalah ....

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Perhatikan bahwa Karena , maka dan . Oleh karena itu, Kemudian, didapat Jadi, jawaban yang tepat adalah B.

Perhatikan bahwa

$begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses x squared plus 4 x minus 5 close parentheses over denominator sin invisible function application open parentheses x squared minus 1 close parentheses end fractionequals limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fractionequals limit as x rightwards arrow 1 of invisible function application open parentheses fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction times fraction numerator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction times fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses end fraction close parenthesesequals limit as x rightwards arrow 1 of invisible function application open parentheses fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction times fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction times fraction numerator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses over denominator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses end fraction close parenthesesequals limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction times limit as x rightwards arrow 1 of invisible function application fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction times limit as x rightwards arrow 1 of invisible function application fraction numerator open parentheses x plus 5 close parentheses up diagonal strike open parentheses x minus 1 close parentheses end strike over denominator open parentheses x plus 1 close parentheses up diagonal strike open parentheses x minus 1 close parentheses end strike end fractionequals limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction times limit as x rightwards arrow 1 of invisible function application fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction times limit as x rightwards arrow 1 of invisible function application fraction numerator open parentheses x plus 5 close parentheses over denominator open parentheses x plus 1 close parentheses end fraction end style$

Karena , maka dan .

Oleh karena itu,

$begin mathsize 14px style limit as x rightwards arrow 1 of invisible function application fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction equals limit as open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses rightwards arrow 0 of invisible function application fraction numerator tan invisible function application open parentheses open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses close parentheses over denominator open parentheses x plus 5 close parentheses open parentheses x minus 1 close parentheses end fraction equals 1limit as x rightwards arrow 1 of invisible function application fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction equals limit as open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses rightwards arrow 0 of invisible function application fraction numerator open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses over denominator sin invisible function application open parentheses open parentheses x plus 1 close parentheses open parentheses x minus 1 close parentheses close parentheses end fraction equals 1 end style$

Kemudian, didapat

Jadi, jawaban yang tepat adalah B.

Buka akses jawaban yang telah terverifikasi