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Pertanyaan

x → − 1 lim ​ 3 x 2 + 8 x + 5 2 x 2 − x − 3 ​

 

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N. Indah

Master Teacher

Mahasiswa/Alumni Universitas Diponegoro

Jawaban terverifikasi

Jawaban

diperoleh .

diperoleh begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 1 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 2 x squared minus x minus 3 over denominator 3 x squared plus 8 x plus 5 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table end style.

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Pembahasan

Dengan cara pemfaktoran dapat ditentukan nilai limit yang diminta sebagai berikut. Dengan demikian, diperoleh .

Dengan cara pemfaktoran dapat ditentukan nilai limit yang diminta sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 1 of space fraction numerator 2 x squared minus x minus 3 over denominator 3 x squared plus 8 x plus 5 end fraction end cell equals cell limit as x rightwards arrow negative 1 of space fraction numerator open parentheses 2 x minus 3 close parentheses open parentheses x plus 1 close parentheses over denominator open parentheses 3 x plus 5 close parentheses open parentheses x plus 1 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow negative 1 of space fraction numerator left parenthesis 2 x minus 3 right parenthesis over denominator left parenthesis 3 x plus 5 right parenthesis end fraction end cell row blank equals cell fraction numerator 2 open parentheses negative 1 close parentheses minus 3 over denominator 3 left parenthesis negative 1 right parenthesis plus 5 end fraction end cell row blank equals cell fraction numerator negative 5 over denominator 2 end fraction end cell row blank equals cell negative 2 1 half end cell end table end style

Dengan demikian, diperoleh begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow negative 1 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 2 x squared minus x minus 3 over denominator 3 x squared plus 8 x plus 5 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table end style.

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muthiah syahidah Humairoh

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