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x → 1 lim ​ x 2 − 2 x + 1 1 − cos ( x − 1 ) ​ = ...

     

  1.  0  

  2.  0 comma 25   

  3. 0 comma 5   

  4.  1    

  5. infinity   

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N. Puspita

Master Teacher

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Ingat kembali sudut rangkap pada cosinus, sifat limitdan limit fungsi trigonometri berikut. Dari aturan di atas, maka diperoleh Dengan demikian, . Jadi, jawaban yang benar adalah C.

Ingat kembali sudut rangkap pada cosinus, sifat limit dan limit fungsi trigonometri berikut.

  • table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 2 x end cell equals cell 1 minus 2 space sin squared x end cell row cell cos space x end cell equals cell 1 minus 2 space sin squared space 1 half x end cell end table  

 

  • limit as x rightwards arrow c of f left parenthesis x right parenthesis times g left parenthesis x right parenthesis equals limit as x rightwards arrow c of space f left parenthesis x right parenthesis times limit as x rightwards arrow c of space g left parenthesis x right parenthesis 

 

  • limit as x rightwards arrow c of k space f left parenthesis x right parenthesis equals k limit as x rightwards arrow c of space f left parenthesis x right parenthesis 

 

  • limit as x rightwards arrow c of fraction numerator sin space a left parenthesis x minus c right parenthesis over denominator b left parenthesis x minus c right parenthesis end fraction equals limit as x rightwards arrow c of fraction numerator space a left parenthesis x minus c right parenthesis over denominator sin space b left parenthesis x minus c right parenthesis end fraction equals a over b  

Dari aturan di atas, maka diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator 1 minus cos space left parenthesis x minus 1 right parenthesis over denominator x squared minus 2 x plus 1 end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus cos space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis squared end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus left parenthesis 1 minus 2 space sin squared space begin display style 1 half end style left parenthesis x minus 1 right parenthesis right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 2 space sin squared space begin display style 1 half end style left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell 2 space limit as x rightwards arrow 1 of space fraction numerator sin space 1 half left parenthesis x minus 1 right parenthesis times sin space 1 half left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell 2 space times space limit as x rightwards arrow 1 of fraction numerator sin space 1 half left parenthesis x minus 1 right parenthesis space over denominator left parenthesis x minus 1 right parenthesis end fraction times limit as x rightwards arrow 1 of fraction numerator sin space 1 half left parenthesis x minus 1 right parenthesis space over denominator left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell 2 times 1 half times 1 half end cell row blank equals cell 1 half end cell row blank equals cell 0 comma 5 end cell end table end style           

Dengan demikian, limit as x rightwards arrow 1 of fraction numerator 1 minus cos space left parenthesis x minus 1 right parenthesis over denominator x squared minus 2 x plus 1 end fraction equals 0 comma 5.

Jadi, jawaban yang benar adalah C.

 

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