x→41​πlim​ 1−tan xcos 2x​= ...

Pertanyaan

limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x over denominator 1 minus tan space x end fraction equals space...         

  1. table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 2 end cell end table    

  2.  table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table  

  3.  table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half square root of 2 end cell end table 

  4. 0 comma 5   

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N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

Pembahasan

Seperti dalam limit fungsi aljabar, di dalam limit fungsi trigonometri, ada limit yang dapat ditentukan nilainya dengan mensubstitusi langsung nilai yang didekati, dan ada yang menghasilkan bentuk tak tentu apabila nilai yang didekati disubstitusi langsung. 

Dengan mengubah bentuk fungsi, maka didapatkan :

limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x over denominator 1 minus tan space x end fraction equals limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x over denominator 1 minus begin display style fraction numerator sin space x over denominator cos space x end fraction end style end fraction equals limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x over denominator begin display style fraction numerator cos space x minus sin space x over denominator cos space x end fraction end style end fraction equals limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x space cos space x over denominator begin display style cos space x minus sin space x end style end fraction equals equals limit as x rightwards arrow 1 fourth straight pi of space fraction numerator open parentheses cos squared space x minus sin squared space x close parentheses space cos space x over denominator begin display style cos space x minus sin space x end style end fraction equals limit as x rightwards arrow 1 fourth straight pi of space fraction numerator up diagonal strike open parentheses cos space x minus sin space x close parentheses end strike open parentheses cos space x plus sin space x close parentheses space cos space x over denominator begin display style up diagonal strike cos space x minus sin space x end strike end style end fraction equals open parentheses cos space 1 fourth straight pi plus sin 1 fourth straight pi close parentheses space cos space 1 fourth straight pi equals open parentheses 1 half square root of 2 plus 1 half square root of 2 close parentheses 1 half square root of 2 equals square root of 2 cross times 1 half square root of 2 equals 1         

Maka, limit as x rightwards arrow 1 fourth straight pi of space fraction numerator cos space 2 x over denominator 1 minus tan space x end fraction equals 1

Oleh karena itu, jawaban yang benar adalah B.

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