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Pertanyaan

Nilai x → 1 lim ​ x − 1 x 2 − 5 x + 4 ​ = ....

  1. -5

  2. -4

  3. -3

  4. 0

  5. 5

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Y. Laksmi

Master Teacher

Mahasiswa/Alumni Universitas Negeri Semarang

Jawaban terverifikasi

Pembahasan

Faktorkan space pembilang  space limit as straight x rightwards arrow 1 of fraction numerator left parenthesis straight x squared minus 5 straight x plus 4 right parenthesis over denominator left parenthesis straight x minus 1 right parenthesis end fraction  equals limit as straight x rightwards arrow 1 of fraction numerator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 4 right parenthesis over denominator left parenthesis straight x minus 1 right parenthesis end fraction  equals limit as straight x rightwards arrow 1 of space open parentheses straight x minus 4 close parentheses    Kemudian space ingat space konsep space limit  limit as straight x rightwards arrow straight a of space straight f left parenthesis straight x right parenthesis space equals space straight f left parenthesis straight a right parenthesis    Pada space konsep space limit space untuk space bentuk space yang space telah space difaktorkan  limit as straight x rightwards arrow 1 of open parentheses space straight x minus 4 close parentheses space equals space 1 minus 4 space equals negative 3

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