Roboguru

x→0lim​(x2sinxsin2x​−x22​)=...

Pertanyaan

limit as x rightwards arrow 0 of left parenthesis fraction numerator sin space 2 x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis equals...      

  1.  negative 4   

  2.  negative 2    

  3. negative 1    

  4.  1    

  5. 16 space    

Pembahasan Video:

Pembahasan:

Ingat kembali sudut rangkap pada sinus dan cosinus, sifat limit dan limit fungsi trigonometri berikut.

  • sin space 2 x equals 2 times sin space x times cos space x

 

  • 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 x end cell row cell cos space x end cell equals cell 1 minus 2 space sin squared space 1 half x end cell end table  

 

  • 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 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 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 0 of fraction numerator sin space a x over denominator b x end fraction equals limit as x rightwards arrow 0 of fraction numerator space a x over denominator sin space b x 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 0 of left parenthesis fraction numerator sin space 2 x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis end cell equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times sin space x times cos space x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times cos space x over denominator x squared end fraction minus 2 over x squared right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of left parenthesis fraction numerator 2 times cos space x minus 2 over denominator x squared end fraction right parenthesis end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis cos space x minus 1 right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis 1 minus 2 space sin squared space begin display style 1 half end style x minus 1 right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of fraction numerator 2 left parenthesis negative 2 space sin squared space begin display style 1 half end style x right parenthesis over denominator x squared end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative 4 times sin space begin display style 1 half end style x times sin space begin display style 1 half end style x over denominator x times x end fraction end cell row blank equals cell negative 4 times limit as x rightwards arrow 0 of space fraction numerator sin space 1 half x over denominator x end fraction times limit as x rightwards arrow 0 of space fraction numerator sin space 1 half x over denominator x end fraction end cell row blank equals cell negative 4 times 1 half times 1 half end cell row blank equals cell negative 1 end cell end table            

Dengan demikian, limit as x rightwards arrow 0 of left parenthesis fraction numerator sin space 2 x over denominator x squared space sin space x end fraction minus 2 over x squared right parenthesis equals negative 1.

Jadi, jawaban yang benar adalah C.

 

Jawaban terverifikasi

Dijawab oleh:

N. Puspita

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan serupa

x→0lim​tan2xsin2xcos2x−1​=...

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved