Pertanyaan

x → 2 lim ​ sin ( 2 x − 4 ) tan ( 3 x − 6 ) ​ = ...

  1.  0

  2.  0 comma 5 

  3. 1 

  4.  1 comma 5  

  5. infinity  

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah D.

jawaban yang benar adalah D.

Pembahasan

Pembahasan

Ingat kembali pemfaktoran,sifat limit dan limit fungsi trigonometri berikut. Dari aturan di atas, maka diperoleh Dengan demikian, . Jadi, jawaban yang benar adalah D.

Ingat kembali pemfaktoran, sifat limit dan limit fungsi trigonometri berikut.

  • a b plus a c equals a left parenthesis b plus c right parenthesis  

 

  •  limit as x rightwards arrow c of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator limit as x rightwards arrow c of space f left parenthesis x right parenthesis over denominator limit as x rightwards arrow c of space g left parenthesis x right parenthesis end fraction  

 

  • 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

 

  • limit as x rightwards arrow c of fraction numerator tan 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 tan space b left parenthesis x minus c right parenthesis 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 2 of fraction numerator tan space left parenthesis 3 x minus 6 right parenthesis over denominator sin space left parenthesis 2 x minus 4 right parenthesis end fraction end cell equals cell limit as x rightwards arrow 2 of space fraction numerator tan space 3 left parenthesis x minus 2 right parenthesis over denominator sin space 2 left parenthesis x minus 2 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 2 of fraction numerator space begin display style fraction numerator tan space 3 left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis end fraction end style over denominator space begin display style fraction numerator sin space 2 left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis end fraction end style end fraction end cell row blank equals cell fraction numerator space limit as x rightwards arrow 2 of begin display style fraction numerator tan space 3 left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis end fraction end style over denominator limit as x rightwards arrow 2 of space begin display style fraction numerator sin space 2 left parenthesis x minus 2 right parenthesis over denominator left parenthesis x minus 2 right parenthesis end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style 3 over 1 end style over denominator begin display style 2 over 1 end style end fraction end cell row blank equals cell 3 over 1 times 1 half end cell row blank equals cell 3 over 2 end cell row blank equals cell 1 comma 5 end cell end table    

Dengan demikian, limit as x rightwards arrow 2 of fraction numerator tan space left parenthesis 3 x minus 6 right parenthesis over denominator sin space left parenthesis 2 x minus 4 right parenthesis end fraction equals 1 comma 5.

Jadi, jawaban yang benar adalah D.

73

Rabiatu Tsani

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Aqila Ramadhania P

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