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Tentukan nilai limit berikut. d. x→1lim​ x2−1(x2−2x+1)(x+2)​

Pertanyaan

Tentukan nilai limit berikut.

d. limit as x rightwards arrow 1 of space fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

nilai dari limit as x rightwards arrow 1 of space fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction adalah 0.

Pembahasan

Ingat mengenai konsep limit as x rightwards arrow c of f open parentheses x close parentheses equals f open parentheses c close parentheses.  

Subtitusikan x rightwards arrow 1 ke persamaan fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction, sehingga didapat:

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

Karena dengan mensubtitusikan  x rightwards arrow 1 ke persamaan fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction didapat 0 over 0, maka fungsi fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction diubah terlebih dahulu menjadi bentuk yang lebih sederhana dengan cara memfaktorkan pembilang dan penyebut.

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

Dengan demikian, nilai dari limit as x rightwards arrow 1 of space fraction numerator open parentheses x squared minus 2 x plus 1 close parentheses open parentheses x plus 2 close parentheses over denominator x squared minus 1 end fraction adalah 0.

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