Pertanyaan

Nilai .... (SIMAK UI 2009)

Nilai ....

(SIMAK UI 2009)

8 dari 10 siswa nilainya naik

dengan paket belajar pilihan

Habis dalam

Klaim

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Pembahasan

Perhatikan perhitungan berikut ini! Kemudian, akan dicari nilai limitnya satu persatu.dengan mengingat Pertama, perhatikan bentuk Didapat . Dengan demikian, diperoleh Selanjutnya, perhatikan bentuk Didapat .Dengan demikian, diperoleh Lalu, perhatikan bentuk Didapat .Dengan demikian, diperoleh Akibatnya, diperoleh hasil sebagai berikut. Jadi, jawabannya adalah C.

Perhatikan perhitungan berikut ini!

Kemudian, akan dicari nilai limitnya satu persatu.dengan mengingat

$begin mathsize 14px style limit as x rightwards arrow infinity of open parentheses square root of a x squared plus b x plus c end root minus square root of p x squared plus q x plus r end root close parentheses equals open curly brackets table attributes columnalign right left end attributes row cell infinity comma end cell cell a greater than p end cell row cell fraction numerator b minus q over denominator 2 square root of a end fraction comma end cell cell a equals p end cell row cell negative infinity comma end cell cell a less than p end cell end table close end style$

Pertama, perhatikan bentuk

Didapat . Dengan demikian, diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of open parentheses square root of 4 x squared plus 8 x end root minus square root of 4 x squared end root close parentheses end cell equals cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row blank equals cell fraction numerator 8 minus 0 over denominator 2 square root of 4 end fraction end cell row blank equals cell fraction numerator 8 over denominator 2 times 2 end fraction end cell row blank equals cell 8 over 4 end cell row blank equals 2 end table end style$

Selanjutnya, perhatikan bentuk

Didapat . Dengan demikian, diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of open parentheses square root of x squared end root minus square root of x squared plus 1 end root close parentheses end cell equals cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row blank equals cell fraction numerator 0 minus 0 over denominator 2 square root of 1 end fraction end cell row blank equals 0 end table end style$

Lalu, perhatikan bentuk

Didapat . Dengan demikian, diperoleh

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of open parentheses square root of x squared end root minus square root of x squared plus x end root close parentheses end cell equals cell fraction numerator b minus q over denominator 2 square root of a end fraction end cell row blank equals cell fraction numerator 0 minus 1 over denominator 2 square root of 1 end fraction end cell row blank equals cell negative 1 half end cell end table end style$

Akibatnya, diperoleh hasil sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow straight infinity of invisible function application open parentheses square root of 4 straight x squared plus 8 straight x end root minus square root of 4 x squared end root close parentheses plus limit as straight x rightwards arrow straight infinity of open parentheses square root of x squared end root minus square root of straight x squared plus 1 end root close parentheses plus limit as straight x rightwards arrow straight infinity of open parentheses square root of x squared end root minus square root of straight x squared plus straight x end root close parentheses end cell row blank equals cell 2 plus 0 plus open parentheses negative 1 half close parentheses end cell row blank equals cell 2 minus 1 half end cell row blank equals cell fraction numerator 4 minus 1 over denominator 2 end fraction end cell row blank equals cell 3 over 2 end cell end table end style$

Jadi, jawabannya adalah C.

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