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Pertanyaan

Tentukan limit-limit fungsi berikut: x → 2 lim ​ x 2 − 3 x + 2 x 2 − 5 x + 6 ​

Tentukan limit-limit fungsi berikut:

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H. Eka

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

nilai

nilai table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared minus 5 x plus 6 over denominator x squared minus 3 x plus 2 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table 

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Pembahasan

Substitusi ke fungsi diperoleh hasil sehingga penyelesaian limit fungsi tersebut menggunakan metode pemfaktoran. Dengan demikian, nilai

Substitusi x equals 2 ke fungsi f open parentheses x close parentheses diperoleh hasil 0 over 0 sehingga penyelesaian limit fungsi tersebut menggunakan metode pemfaktoran.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of space fraction numerator x squared minus 5 x plus 6 over denominator x squared minus 3 x plus 2 end fraction end cell equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses x minus 3 close parentheses open parentheses x minus 2 close parentheses over denominator open parentheses x minus 1 close parentheses open parentheses x minus 2 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 2 of space fraction numerator open parentheses x minus 3 close parentheses over denominator open parentheses x minus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 2 minus 3 over denominator 2 minus 1 end fraction end cell row blank equals cell fraction numerator negative 1 over denominator 1 end fraction end cell row blank equals cell negative 1 end cell end table

Dengan demikian, nilai table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared minus 5 x plus 6 over denominator x squared minus 3 x plus 2 end fraction end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table 

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Selidiki fungsi tersebut mempunyai limit atau tidak x → 1 lim ​ x 2 − 1 x 4 − 1 ​ ,berikan alasan!

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