Roboguru

Nilai dari  adalah ....

Pertanyaan

Nilai dari limit as x rightwards arrow 0 of space fraction numerator 1 minus cos space x over denominator x space sin space 2 x end fraction adalah ....

  1. 0 

  2. 1 over 8 

  3. 1 fourth 

  4. 1 half 

  5. 1 

Pembahasan Soal:

Dapat ditentukan nilai limit yang diminta sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator 1 minus cos space x over denominator x space sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator 1 minus cos space x over denominator x space sin space 2 x end fraction times fraction numerator 1 plus cos space x over denominator 1 plus space cos space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator 1 minus cos squared space x over denominator x space open parentheses 2 times sin space x times cos space x close parentheses open parentheses 1 plus cos space x close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator sin squared space x over denominator 2 x times sin space x times cos space x left parenthesis 1 plus cos space x right parenthesis end fraction end cell row blank equals cell space limit as x rightwards arrow 0 of space fraction numerator sin space x over denominator x end fraction times limit as x rightwards arrow 0 of space fraction numerator up diagonal strike sin space x end strike over denominator 2 up diagonal strike sin space x end strike times cos space x left parenthesis 1 plus cos space x right parenthesis end fraction end cell row blank equals cell 1 times limit as x rightwards arrow 0 of space fraction numerator 1 over denominator 2 space cos space x left parenthesis 1 plus cos space x right parenthesis end fraction end cell row blank equals cell 1 times fraction numerator 1 over denominator 2 times cos space 0 times open parentheses 1 plus cos space 0 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 times 1 times open parentheses 1 plus 1 close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator 2 times 1 times 2 end fraction end cell row blank equals cell 1 fourth end cell end table

Jadi, jawaban yang benar adalah C. 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 13 Juni 2021

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