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x → 0 lim ​ ( x 2 sin x sin 2 x ​ − x 2 2 ​ ) = ...

      

  1.  negative 4   

  2.  negative 2    

  3. negative 1    

  4.  1    

  5. 16 space    

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N. Puspita

Master Teacher

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

  • sin space 2 x equals 2 times sin space x times cos space x

 

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

 

  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator b x end fraction equals limit as x rightwards arrow 0 of fraction numerator space a x over denominator sin space b x end fraction equals a over b  

Dari aturan di atas, maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of left parenthesis fraction numerator sin space 2 x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis end cell equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times sin space x times cos space x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times cos space x over denominator x squared end fraction minus 2 over x squared right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times cos space x minus 2 over denominator x squared end fraction right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis cos space x minus 1 right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis 1 minus 2 space sin squared space begin display style 1 half end style x minus 1 right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis negative 2 space sin squared space begin display style 1 half end style x right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative 4 times sin space begin display style 1 half end style x times sin space begin display style 1 half end style x over denominator x times x end fraction end cell row blank equals cell negative 4 times limit as x rightwards arrow 0 of space fraction numerator sin space 1 half x over denominator x end fraction times limit as x rightwards arrow 0 of space fraction numerator sin space 1 half x over denominator x end fraction end cell row blank equals cell negative 4 times 1 half times 1 half end cell row blank equals cell negative 1 end cell end table            

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

Jadi, jawaban yang benar adalah C.

 

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