Pertanyaan

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

$x \to 3 lim \frac{x ^{2} - 9}{x + 1 - 7 - x} =$ ....

(SBMPTN 2018)

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Y. Endah

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Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

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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