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Pertanyaan

x → − 1 lim ​ x 2 + 4 x + 3 x 2 − 1 ​ = ...

 ...

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S. Ayu

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Prof. DR. Hamka

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Pembahasan

Dengan menggunakan metode pemfaktoran, diperoleh: Maka, hasil dari .

Dengan menggunakan metode pemfaktoran, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 1 of space fraction numerator x squared minus 1 over denominator x squared plus 4 x plus 3 end fraction end cell equals cell limit as x rightwards arrow negative 1 of space fraction numerator x squared minus x plus x minus 1 over denominator x squared plus 3 x plus x plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of space fraction numerator x open parentheses x minus 1 close parentheses plus 1 open parentheses x minus 1 close parentheses over denominator x open parentheses x plus 3 close parentheses plus 1 open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of fraction numerator down diagonal strike open parentheses x plus 1 close parentheses end strike open parentheses x minus 1 close parentheses over denominator down diagonal strike open parentheses x plus 1 close parentheses end strike open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of fraction numerator x minus 1 over denominator x plus 3 end fraction end cell row blank equals cell fraction numerator negative 1 minus 1 over denominator negative 1 plus 3 end fraction end cell row blank equals cell fraction numerator negative 2 over denominator 2 end fraction end cell row blank equals cell negative 1 end cell end table

Maka, hasil dari limit as x rightwards arrow negative 1 of space fraction numerator x squared minus 1 over denominator x squared plus 4 x plus 3 end fraction equals negative 1.

46

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