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Pertanyaan

Nilai x → 1 lim ​ x − 1 2 x 2 − 7 x + 5 ​ = ....

Nilai

  1. -5

  2. -3

  3. 4

  4. 5

  5. 10

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R. Diah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

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Pembahasan

Dengan cara difaktorkan

Dengan cara difaktorkan

limit as straight x rightwards arrow 1 of fraction numerator 2 straight x squared minus 7 straight x plus 5 over denominator straight x minus 1 end fraction  limit as straight x rightwards arrow 1 of fraction numerator left parenthesis 2 straight x minus 5 right parenthesis left parenthesis straight x minus 1 over denominator left parenthesis straight x minus 1 right parenthesis end fraction  limit as straight x rightwards arrow 1 of 2 straight x minus 5  space space space space space space space equals 2 left parenthesis 1 right parenthesis minus 5  space space space space space space space equals negative 3

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Pertanyaan serupa

x → − 5 lim ​ 4 x 2 + 2 x − 1 = ...

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Jawaban terverifikasi

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