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Pertanyaan

x → 3 lim ​ x + 1 ​ − 7 − x ​ x 2 − 9 ​ = .... (SBMPTN 2018)

....

(SBMPTN 2018)

  1. begin mathsize 14px style 8 end style

  2. begin mathsize 14px style 12 end style

  3. begin mathsize 14px style 16 end style

  4. begin mathsize 14px style 20 end style

  5. begin mathsize 14px style 24 end style

8 dari 10 siswa nilainya naik

dengan paket belajar pilihan

Habis dalam

01

:

20

:

42

:

02

Klaim

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Y. Endah

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

jawabannya adalah B.

jawabannya adalah B.

Pembahasan

Perhatikan bahwa Jadi, jawabannya adalah B.

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of space fraction numerator x squared minus 9 over denominator square root of x plus 1 end root minus square root of 7 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses over denominator square root of x plus 1 end root minus square root of 7 minus x end root end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space open parentheses fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses over denominator square root of x plus 1 end root minus square root of 7 minus x end root end fraction times fraction numerator square root of x plus 1 end root plus square root of 7 minus x end root over denominator square root of x plus 1 end root plus square root of 7 minus x end root end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses open parentheses square root of x plus 1 end root plus square root of 7 minus x end root close parentheses over denominator x plus 1 minus open parentheses 7 minus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses open parentheses square root of x plus 1 end root plus square root of 7 minus x end root close parentheses over denominator x plus 1 minus 7 plus x end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses open parentheses square root of x plus 1 end root plus square root of 7 minus x end root close parentheses over denominator 2 x minus 6 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses open parentheses square root of x plus 1 end root plus square root of 7 minus x end root close parentheses over denominator 2 open parentheses x minus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of space fraction numerator open parentheses x plus 3 close parentheses open parentheses square root of x plus 1 end root plus square root of 7 minus x end root close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator open parentheses 3 plus 3 close parentheses open parentheses square root of 3 plus 1 end root plus square root of 7 minus 3 end root close parentheses over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 open parentheses square root of 4 plus square root of 4 close parentheses over denominator 2 end fraction end cell row blank equals cell 3 times left parenthesis 2 plus 2 right parenthesis end cell row blank equals cell 3 times 4 end cell row blank equals 12 end table end style

Jadi, jawabannya adalah B.

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