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Jika , maka hasil dari adalah ...

Jika begin mathsize 14px style f open parentheses x close parentheses equals 2 x squared plus x minus 1 end style, maka hasil dari begin mathsize 14px style limit as p rightwards arrow 0 of space fraction numerator f open parentheses x plus p close parentheses minus f open parentheses x close parentheses over denominator p end fraction end style adalah ...

  1. begin mathsize 14px style 4 x end style 

  2. size 14px 4 size 14px x size 14px plus size 14px 1 

  3. size 14px 4 size 14px x size 14px minus size 14px 1 

  4. size 14px x 

  5. begin mathsize 14px style 1 half x end style 

Jawaban:

Dengan menerapkan sifat-sifat limit fungsi, diperoleh perhitungan sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as p rightwards arrow 0 of space fraction numerator f open parentheses x plus p close parentheses minus f open parentheses x close parentheses over denominator p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space fraction numerator open parentheses 2 open parentheses x plus p close parentheses squared plus open parentheses x plus p close parentheses minus 1 close parentheses minus open parentheses 2 x squared plus x minus 1 close parentheses over denominator p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space fraction numerator open parentheses 2 open parentheses x squared plus 2 x p plus p squared close parentheses plus x plus p minus 1 close parentheses minus open parentheses 2 x squared plus x minus 1 close parentheses over denominator p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space fraction numerator open parentheses up diagonal strike 2 x squared end strike plus 4 x p plus 2 p squared plus up diagonal strike x plus p minus up diagonal strike 1 close parentheses minus open parentheses up diagonal strike 2 x squared end strike plus up diagonal strike x minus up diagonal strike 1 close parentheses over denominator p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space fraction numerator 4 x p plus 2 p squared plus p over denominator p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space fraction numerator up diagonal strike p open parentheses 4 x plus 2 p plus 1 close parentheses over denominator up diagonal strike p end fraction end cell row blank equals cell limit as p rightwards arrow 0 of space open parentheses 4 x plus 2 p plus 1 close parentheses end cell row blank equals cell 4 x plus 2 open parentheses 0 close parentheses plus 1 end cell row blank equals cell 4 x plus 1 end cell end table end style 

Jadi, hasil dari begin mathsize 14px style limit as p rightwards arrow 0 of space fraction numerator f open parentheses x plus p close parentheses minus f open parentheses x close parentheses over denominator p end fraction end style adalah size 14px 4 size 14px x size 14px plus size 14px 1.

Dengan demikian, jawaban yang tepat adalah B.

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