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Pertanyaan

x → 3 lim ​ x + 1 ​ − 7 − x ​ x 2 − 9 ​ = ....

  1. 8

  2. 12

  3. 16

  4. 20

  5. 24

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D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

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Pembahasan

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of invisible function application 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 equals cell limit as x rightwards arrow 3 of invisible function application 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. 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 end cell row blank equals cell limit as x rightwards arrow 3 of invisible function application fraction numerator open parentheses x squared minus 9 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 open parentheses square root of x plus 1 end root close parentheses squared minus open parentheses square root of 7 minus x end root close parentheses squared end fraction end cell row blank equals cell limit as x rightwards arrow 3 of invisible function application fraction numerator open parentheses x squared minus 9 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 open parentheses x plus 1 close parentheses minus open parentheses 7 minus x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 3 of invisible function application fraction numerator open parentheses x squared minus 9 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 invisible function application 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 invisible function application 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 24 over 2 end cell row blank equals cell box enclose 12 end cell end table

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Pertanyaan serupa

Tentukan nilai dari x → − 2 lim ​ x + 2 2 x + 13 ​ − x + 11 ​ ​ .

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