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Jika dan , maka nilai dari x → ∞ lim ​ det ( A ) det ( B ) ​ adalah ....

Jika begin mathsize 14px style straight A equals open square brackets table row 0 cell cotan blank open parentheses 1 over x close parentheses end cell row cell negative 1 end cell 0 end table close square brackets end style dan begin mathsize 14px style straight B equals open square brackets table row cell 1 over x squared cosec blank open parentheses 1 over x close parentheses space end cell 0 row 0 x end table close square brackets end style,

maka nilai dari  adalah .... 

  1. begin mathsize 14px style negative infinity end style undefined 

  2. begin mathsize 14px style negative 1 end style undefined 

  3. begin mathsize 14px style 0 end style undefined 

  4. begin mathsize 14px style 1 end style undefined 

  5. begin mathsize 14px style infinity end style undefined 

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F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

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Pembahasan

Diketahui maka didapat Selanjutnya diketahui maka didapat Sehingga didapat Misalkan . Jika maka . Oleh karena itu didapat Dengan demikian, nilai dari adalah 0. Jadi, jawaban yang tepat adalah C. Untuk mempelajarinya lebih jelas, tonton video selanjutnya.

Diketahui begin mathsize 14px style straight A equals open square brackets table row 0 cell cotan open parentheses 1 over x close parentheses end cell row cell negative 1 end cell 0 end table close square brackets end style 

maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell det left parenthesis straight A right parenthesis end cell equals cell 0 times 0 minus cotan open parentheses 1 over x close parentheses times left parenthesis negative 1 right parenthesis end cell row blank equals cell 0 plus cotan open parentheses 1 over x close parentheses end cell row blank equals cell cotan open parentheses 1 over x close parentheses end cell end table end style  

Selanjutnya diketahui begin mathsize 14px style straight B equals open square brackets table row cell 1 over x squared cosec open parentheses 1 over x close parentheses end cell 0 row 0 x end table close square brackets end style 

maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell det left parenthesis straight B right parenthesis end cell equals cell 1 over x squared cosec open parentheses 1 over straight x close parentheses times x minus 0 times 0 end cell row blank equals cell 1 over x cosec open parentheses 1 over straight x close parentheses minus 0 end cell row blank equals cell 1 over x cosec open parentheses 1 over straight x close parentheses end cell end table end style 

Sehingga didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator det left parenthesis straight B right parenthesis over denominator det left parenthesis straight A right parenthesis end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator begin display style 1 over straight x end style cosec open parentheses begin display style 1 over x end style close parentheses over denominator cotan open parentheses begin display style 1 over straight x end style close parentheses end fraction end cell end table end style 

Misalkan begin mathsize 14px style y equals 1 over x end style.

Jika begin mathsize 14px style x rightwards arrow infinity end style maka begin mathsize 14px style y rightwards arrow 0 end style.

Oleh karena itu didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow infinity of fraction numerator det left parenthesis straight B right parenthesis over denominator det left parenthesis straight A right parenthesis end fraction end cell equals cell limit as x rightwards arrow infinity of fraction numerator begin display style 1 over x end style cosec open parentheses begin display style 1 over x end style close parentheses over denominator cotan open parentheses begin display style 1 over x end style close parentheses end fraction end cell row blank equals cell limit as y rightwards arrow 0 of fraction numerator begin display style y space cosec open parentheses y close parentheses end style over denominator cotan open parentheses y close parentheses end fraction end cell row blank equals cell limit as y rightwards arrow 0 of fraction numerator begin display style y times fraction numerator 1 over denominator sin open parentheses straight y close parentheses end fraction end style over denominator begin display style fraction numerator cos left parenthesis y right parenthesis over denominator sin left parenthesis y right parenthesis end fraction end style end fraction end cell row blank equals cell limit as y rightwards arrow 0 of open parentheses y times fraction numerator 1 over denominator sin open parentheses straight y close parentheses end fraction times fraction numerator sin left parenthesis y right parenthesis over denominator cos left parenthesis y right parenthesis end fraction close parentheses end cell row blank equals cell limit as y rightwards arrow 0 of open parentheses fraction numerator y over denominator cos left parenthesis y right parenthesis end fraction close parentheses end cell row blank equals cell fraction numerator 0 over denominator cos left parenthesis 0 right parenthesis end fraction end cell row blank equals cell 0 over 1 end cell row blank equals 0 end table end style 

Dengan demikian, nilai dari begin mathsize 14px style limit as x rightwards arrow infinity of invisible function application fraction numerator det invisible function application straight B over denominator det invisible function application straight A end fraction end style adalah 0.

Jadi, jawaban yang tepat adalah C.undefined 

Untuk mempelajarinya lebih jelas, tonton video selanjutnya.

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