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x→0lim​ (3−9−x​2x​)=....

Pertanyaan

limit as x rightwards arrow 0 of space open parentheses fraction numerator 2 x over denominator 3 minus square root of 9 minus x end root end fraction close parentheses equals....

  1. 18

  2. 15

  3. 12

  4. 9

  5. 6

T. Prita

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jember

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Dalam mengerjakan limit, jika substitusi langsung menghasilkan bentuk tak tentu maka harus menggunakan cara lain dalam mencari nilainya menggunakan operasi aljabar misalnya memfaktorkan, membagi, merasionalkan, dan lainnya.

Beberapa bentuk tak tentu antara lain: 0 over 0 comma space infinity over infinity comma space open parentheses infinity minus infinity close parentheses comma space open parentheses 0 times infinity close parentheses

Substitusi nilai x equals 0 diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space open parentheses fraction numerator 2 x over denominator 3 minus square root of 9 minus x end root end fraction close parentheses end cell equals cell fraction numerator 2 open parentheses 0 close parentheses over denominator 3 minus square root of 9 minus 0 end root end fraction end cell row blank equals cell fraction numerator 0 over denominator 3 minus square root of 9 end fraction end cell row blank equals cell fraction numerator 0 over denominator 3 minus 3 end fraction end cell row blank equals cell 0 over 0 space open parentheses text bentuk tak tentu end text close parentheses end cell end table

Selanjutnya dengan mengkalikan akar sekawan diperoleh:

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

Oleh karena itu, jawaban yang benar adalah C.

157

5.0 (15 rating)

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