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x→4lim​ (x2−4xx+4​−x−42​)=....

Pertanyaan

limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses equals....

  1. negative 1

  2. negative 0 comma 5

  3. negative 0 comma 25

  4. 0

  5. 0 comma 25

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.

Diketahui: 

limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses, dapat dituliskan:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell equals cell limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x open parentheses x minus 4 close parentheses end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator x plus 4 minus 2 open parentheses x close parentheses over denominator x open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator x plus 4 minus 2 x over denominator x open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator negative x plus 4 over denominator x open parentheses x minus 4 close parentheses end fraction end cell end table

Sehingga table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell equals cell limit as x rightwards arrow 4 of space fraction numerator negative x plus 4 over denominator x open parentheses x minus 4 close parentheses end fraction end cell end table

Substitusi nilai x equals 4 diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell equals cell limit as x rightwards arrow 4 of space fraction numerator negative x plus 4 over denominator x open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell fraction numerator negative 4 plus 4 over denominator 4 open parentheses 4 minus 4 close parentheses end fraction end cell row blank equals cell fraction numerator negative 4 plus 4 over denominator 4 open parentheses 0 close parentheses 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 memfaktorkan diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space open parentheses fraction numerator x plus 4 over denominator x squared minus 4 x end fraction minus fraction numerator 2 over denominator x minus 4 end fraction close parentheses end cell equals cell limit as x rightwards arrow 4 of space fraction numerator negative x plus 4 over denominator x open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator negative open parentheses x minus 4 close parentheses over denominator x open parentheses x minus 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator negative 1 over denominator x end fraction end cell row blank equals cell negative 1 fourth end cell row blank equals cell negative 0 comma 25 end cell end table

Oleh karena itu, jawaban yang benar adalah C.

86

5.0 (1 rating)

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