x→1lim(x−12−x2−x2)=....

Pertanyaan

$limit as x rightwards arrow 1 of space open parentheses fraction numerator 2 over denominator x minus 1 end fraction minus fraction numerator 2 over denominator x squared minus x end fraction close parentheses equals....$

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: $\frac00,\;\frac\infty\infty,\;\left(\infty-\infty\right),\;\left(0\cdot\infty\right)$ .

Diketahui:

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

Sehingga $table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space open parentheses fraction numerator 2 over denominator x minus 1 end fraction minus fraction numerator 2 over denominator x squared minus x end fraction close parentheses end cell equals cell limit as x rightwards arrow 1 of space fraction numerator 2 x minus 2 over denominator x open parentheses x minus 1 close parentheses end fraction end cell end table$

Substitusi nilai $x=1$ diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator 2 x minus 2 over denominator x open parentheses x minus 1 close parentheses end fraction end cell equals cell fraction numerator 2 open parentheses 1 close parentheses minus 2 over denominator 1 open parentheses 1 minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 2 minus 2 over denominator 1 open parentheses 0 close parentheses end fraction end cell row blank equals cell 0 over 0 space open parentheses text bentuk tak tentu end text close parentheses end cell end table$

Selanjutnya dengan memfaktorkan diperoleh:

Oleh karena itu, jawaban yang benar adalah C.