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Pertanyaan

Nilai x → 5 lim ​ 2 x 2 + x − 3 2 x + 3 ​ = …

Nilai  

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R. Almira

Master Teacher

Mahasiswa/Alumni Universitas Pertamina

Jawaban terverifikasi

Jawaban

nilai .

nilai begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator 2 x squared plus x minus 3 end fraction equals 1 fourth end style

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Pembahasan

Dengan menggunakan metode pemfaktoran dan substitusi, nilai limit tersebut yaitu: Jadi, nilai .

Dengan menggunakan metode pemfaktoran dan substitusi, nilai limit tersebut yaitu:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator 2 x squared plus x minus 3 end fraction end cell equals cell limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator open parentheses 2 x plus 3 close parentheses open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 5 of fraction numerator 1 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 1 over denominator left parenthesis 5 right parenthesis minus 1 end fraction end cell row blank equals cell 1 fourth end cell end table end style 

Jadi, nilai begin mathsize 14px style limit as x rightwards arrow 5 of fraction numerator 2 x plus 3 over denominator 2 x squared plus x minus 3 end fraction equals 1 fourth end style

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