Roboguru

Nilai x→0lim​tan2x2x2+4x​=....

Pertanyaan

Nilai stack l i m with x rightwards arrow 0 below space fraction numerator 2 x squared space plus space 4 x over denominator tan begin display style space end style begin display style 2 end style begin display style x end style end fraction space equals space. space. space. space.

  1. 0

  2. 1

  3. 2

  4. 3

  5. 4

Pembahasan Soal:

Jawaban: C

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Endah

Mahasiswa/Alumni Institut Teknologi Bandung

Terakhir diupdate 04 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

x→1lim​(x−1)tan(3x−3)cos(2x−2)−1​=...

Pembahasan Soal:

Ingat kembali beberapa sifat berikut.

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

 

  • 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 space x end cell end table  

 

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

 

  • limit as x rightwards arrow c of fraction numerator sin space a left parenthesis x minus 1 right parenthesis over denominator b left parenthesis x minus 1 right parenthesis end fraction equals limit as x rightwards arrow 0 of fraction numerator space a left parenthesis x minus 1 right parenthesis over denominator sin space b left parenthesis x minus 1 right parenthesis end fraction equals a over b

 

  • limit as x rightwards arrow c of fraction numerator sin space a left parenthesis x minus 1 right parenthesis over denominator tan space b left parenthesis x minus 1 right parenthesis end fraction equals a over b     

Dari aturan di atas, maka diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of fraction numerator cos space left parenthesis 2 x minus 2 right parenthesis minus 1 over denominator left parenthesis x minus 1 right parenthesis space tan space left parenthesis 3 x minus 3 right parenthesis end fraction end cell equals cell limit as x rightwards arrow 1 of fraction numerator cos space 2 left parenthesis x minus 1 right parenthesis minus 1 over denominator left parenthesis x minus 1 right parenthesis space tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator 1 minus 2 space sin squared space left parenthesis x minus 1 right parenthesis minus 1 over denominator left parenthesis x minus 1 right parenthesis space tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator negative 2 space sin squared space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis space tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 of fraction numerator negative 2 space sin space left parenthesis x minus 1 right parenthesis space sin space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis space tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell negative 2 limit as x rightwards arrow 1 of fraction numerator sin space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis end fraction times fraction numerator sin space left parenthesis x minus 1 right parenthesis over denominator tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell negative 2 times limit as x rightwards arrow 1 of fraction numerator sin space left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis end fraction limit as x rightwards arrow 1 of fraction numerator sin space left parenthesis x minus 1 right parenthesis over denominator tan space 3 left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell negative 2 times 1 times 1 third end cell row blank equals cell negative 2 over 3 end cell end table end style   

 

Dengan demikian, limit as x rightwards arrow 1 of fraction numerator cos space left parenthesis 2 x minus 2 right parenthesis minus 1 over denominator left parenthesis x minus 1 right parenthesis space tan space left parenthesis 3 x minus 3 right parenthesis end fraction equals negative 2 over 3.

Jadi, jawaban yang benar adalah A.

 

0

Roboguru

Tentukanlah h→0lim​htan(x+h)−tanx​

Pembahasan Soal:

Ingat kembali rumus selisih tangen, sifat limit dan limit fungsi trigonometri berikut.

  • tan space A minus tan space B equals tan space left parenthesis A minus B right parenthesis left parenthesis 1 plus tan space A space tan space B right parenthesis  

 

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

 

Dari aturan di atas, maka diperoleh

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as h rightwards arrow 0 of fraction numerator tan left parenthesis x plus h right parenthesis minus tan space x over denominator h end fraction end cell equals cell limit as h rightwards arrow 0 of fraction numerator tan space left parenthesis x plus h minus x right parenthesis left parenthesis 1 plus tan left parenthesis x plus h right parenthesis space tan space x right parenthesis over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator tan space h left parenthesis 1 plus tan left parenthesis x plus h right parenthesis space tan space x right parenthesis over denominator h end fraction end cell row blank equals cell limit as h rightwards arrow 0 of fraction numerator tan space h over denominator h end fraction times limit as h rightwards arrow 0 of left parenthesis 1 plus tan left parenthesis x plus h right parenthesis space tan space x right parenthesis end cell row blank equals cell 1 times left parenthesis 1 plus tan left parenthesis x plus 0 right parenthesis tan space x right parenthesis end cell row blank equals cell 1 plus tan space x space tan space x end cell row blank equals cell 1 plus tan squared x end cell end table end style 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row cell limit as h rightwards arrow 0 of fraction numerator tan left parenthesis x plus h right parenthesis minus tan space x over denominator h end fraction end cell equals cell 1 plus tan squared x end cell end table

1

Roboguru

x→0lim​xtanxcosx−cos2x​=...

Pembahasan Soal:

Ingatlah rumus selisih cosinus

cos space straight A minus cos space straight B equals negative 2 space sin space 1 half open parentheses straight A plus straight B close parentheses space sin space 1 half open parentheses straight A minus straight B close parentheses

serta sifat-sifat limit fungsi

  • limit as x rightwards arrow c of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow c of f left parenthesis x right parenthesis
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis times g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis times limit as x rightwards arrow c of g 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 a over b
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b

Dengan menggunakan sifat-sifat limit fungsi dan rumus selisih cosinus tersebut, persoalan limit di atas dapat diselesaikan dengan cara sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses cos space 2 x minus cos space x close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative open parentheses negative 2 space sin 1 half open parentheses 2 x plus x close parentheses space sin 1 half open parentheses 2 x minus x close parentheses close parentheses over denominator x space tan space x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator 2 space sin space begin display style 3 over 2 end style x space sin space 1 half x over denominator x space tan space x end fraction end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space open parentheses fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction close parentheses end cell row blank equals cell 2 times limit as x rightwards arrow 0 of space fraction numerator sin space begin display style 3 over 2 end style x over denominator x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space begin display style 1 half end style x over denominator tan space x end fraction end cell row blank equals cell 2 times open parentheses fraction numerator begin display style 3 over 2 end style over denominator 1 end fraction close parentheses times open parentheses fraction numerator begin display style 1 half end style over denominator 1 end fraction close parentheses end cell row blank equals cell 3 over 2 end cell row blank equals cell 1 comma 5 end cell end table  

Maka, limit as x rightwards arrow 0 of space fraction numerator cos space x minus cos space 2 x over denominator x space tan space x end fraction equals 1 comma 5

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Nilai x→0lim​2sin2x−tanxtan6x​ adalah ....

Pembahasan Soal:

Rumus limit trigonometri untuk x rightwards arrow 0 yang didapat dari penurunan dengan kaidah L'Hopital sebagai berikut.

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

Selain itu, ingatlah identitas trigonometri berikut.

left parenthesis straight i right parenthesis space tan space x equals fraction numerator sin space x over denominator cos space x end fraction left parenthesis ii right parenthesis space sin space 2 x equals 2 space sin space x space cos space x

Pertama, lakukan metode substitusi untuk menjawab soal limit di atas.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator tan space 6 x over denominator 2 space sin space 2 x minus tan space x end fraction end cell equals cell fraction numerator tan space 6 left parenthesis 0 right parenthesis over denominator 2 space sin space 2 left parenthesis 0 right parenthesis minus tan space left parenthesis 0 right parenthesis end fraction end cell row blank equals cell fraction numerator 0 over denominator 2 left parenthesis 0 right parenthesis minus 0 end fraction end cell row blank equals cell 0 over 0 space left parenthesis bentuk space tak space tentu right parenthesis end cell end table

Karena menghasilkan bentuk tak tentu, penyelesaian limit di atas adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator tan space 6 x over denominator 2 space sin space 2 x minus tan space x end fraction end cell equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator 1 over denominator 2 space sin space 2 x minus begin display style fraction numerator sin space x over denominator cos space x end fraction end style end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator 1 over denominator 2 space sin space 2 x minus begin display style fraction numerator sin space x over denominator cos space x end fraction end style end fraction cross times fraction numerator cos space x over denominator cos space x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator cos space x over denominator 2 space sin space 2 x space cos space x minus sin space x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator cos space x over denominator 2 times 2 space sin space x times cos space x times cos space x minus sin space x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator cos space x over denominator 4 space sin space x times cos squared x minus sin space x end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator cos space 6 x end fraction times fraction numerator cos space x over denominator sin space x open parentheses 4 space cos squared x minus 1 close parentheses end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator sin space x end fraction times fraction numerator cos space x over denominator cos space 6 x end fraction times fraction numerator 1 over denominator 4 space cos squared x minus 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 0 of open parentheses fraction numerator sin space 6 x over denominator sin space x end fraction close parentheses times limit as x rightwards arrow 0 of open parentheses fraction numerator cos space x over denominator cos space 6 x end fraction close parentheses times limit as x rightwards arrow 0 of open parentheses fraction numerator 1 over denominator 4 space cos squared x minus 1 end fraction close parentheses end cell row blank equals cell 6 over 1 times fraction numerator cos space 0 over denominator cos space 6 open parentheses 0 close parentheses end fraction times fraction numerator 1 over denominator 4 open parentheses cos open parentheses 0 close parentheses close parentheses squared minus 1 end fraction end cell row blank equals cell 6 times 1 over 1 times fraction numerator 1 over denominator 4 minus 1 end fraction end cell row blank equals cell 6 times 1 times 1 third end cell row blank equals cell 6 over 3 end cell row blank equals 2 end table

Dengan demikian, hasil dari limit as x rightwards arrow 0 of fraction numerator tan space 6 x over denominator 2 space sin space 2 x minus tan space x end fraction adalah 2.

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

x→0lim​tan2xsin2xcos2x−1​=...

Pembahasan Soal:

Ingatlah identitas cosinus sudut rangkap

cos space 2 x equals 1 minus 2 space sin squared space x

serta sifat-sifat limit fungsi

  • limit as x rightwards arrow c of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow c of f left parenthesis x right parenthesis
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis times g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis times limit as x rightwards arrow c of g left parenthesis x right parenthesis
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b

Dengan menggunakan sifat-sifat limit fungsi dan identitas tersebut, didapatkan :

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses 1 minus 2 sin to the power of 2 space end exponent x close parentheses minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative 2 sin to the power of 2 space end exponent x over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator tan space 2 x end fraction times fraction numerator sin space x over denominator sin space 2 x end fraction close parentheses end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator tan space 2 x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator sin space 2 x end fraction end cell row blank equals cell negative 2 times 1 half times 1 half end cell row blank equals cell fraction numerator negative 2 over denominator 4 end fraction end cell row blank equals cell negative 0 comma 5 end cell end table   

Maka, limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction equals negative 0 comma 5

Oleh karena itu, jawaban yang benar adalah B.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

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