Roboguru

Diketahui ,

Pertanyaan

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell x squared minus 1 space untuk space x less than negative 2 end cell row cell 2 x minus 1 space untuk space x greater than negative 2 end cell end table close end style,

begin mathsize 14px style limit as x rightwards arrow negative 2 of f left parenthesis x right parenthesis equals.... end style   

  1. ...undefined 

  2. ...undefined 

Pembahasan Video:

Pembahasan Soal:

Terlebih dahulu, kita cari limit kiri dan limit kanan.

begin mathsize 14px style f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell x squared minus 1 space untuk space x less than negative 2 end cell row cell 2 x minus 1 space untuk space x greater than negative 2 end cell end table close end style

Limit kiri:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 to the power of minus of f left parenthesis x right parenthesis end cell equals cell left parenthesis negative 2 right parenthesis squared minus 1 end cell row blank equals 3 end table end style

Limit kanan:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 to the power of plus of f left parenthesis x right parenthesis end cell equals cell 2 left parenthesis negative 2 right parenthesis minus 1 end cell row blank equals cell negative 5 end cell end table end style

Karena begin mathsize 14px style limit space kiri not equal to limit space kanan end style, maka begin mathsize 14px style limit as x rightwards arrow negative 2 of f left parenthesis x right parenthesis end style tidak ada.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Freelancer6

Terakhir diupdate 03 Mei 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 yang serupa

Hitunglah setiap limit fungsi berikut. a.   b.

Pembahasan Soal:

Ingat bahwa:

Nilai limit suatu fungsi ada jika limit kanan dan limit kirinya sama.

a. limit as x rightwards arrow 2 of f left parenthesis x right parenthesis comma space untuk space f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell 3 minus 2 x end cell cell x less or equal than 2 end cell row cell x squared minus 5 end cell cell x greater than 2 end cell end table close

Berdasarkan soal di atas dapat diketahui bahwa

1) Limit kanannya adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 to the power of plus of space f left parenthesis x right parenthesis end cell equals cell limit as x rightwards arrow 2 to the power of plus of space 3 minus 2 x end cell row blank equals cell 3 minus 2 left parenthesis 2 right parenthesis end cell row blank equals cell 3 minus 4 end cell row blank equals cell negative 1 end cell end table

2) Limit kirinya adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 to the power of minus of f left parenthesis x right parenthesis end cell equals cell limit as x rightwards arrow 2 to the power of minus of x squared minus 5 end cell row blank equals cell left parenthesis 2 right parenthesis squared minus 5 end cell row blank equals cell 4 minus 5 end cell row blank equals cell negative 1 end cell end table

Sehingga diperoleh limit kanan dan limit kirinya adalah negative 1. Karena limit kanan dan kirinya sama maka fungsi di atas memiliki limit yaitu negative 1.

Dengan demikian, nilai limit dari limit as x rightwards arrow 2 of f left parenthesis x right parenthesis comma space untuk space f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell 3 minus 2 x end cell cell x less or equal than 2 end cell row cell x squared minus 5 end cell cell x greater than 2 end cell end table close adalah negative 1.

b. limit as x rightwards arrow 1 of g left parenthesis x right parenthesis comma space untuk space g left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator x squared minus x over denominator x minus 1 end fraction end cell cell x less than 1 end cell row cell square root of 1 minus x end root end cell cell x greater or equal than 1 end cell end table close

Berdasarkan soal di atas dapat diketahui bahwa

1) Limit kanannya adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 to the power of plus of g left parenthesis x right parenthesis end cell equals cell limit as x rightwards arrow 1 to the power of plus of fraction numerator x squared minus x over denominator x minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 to the power of plus of fraction numerator x left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis end fraction end cell row blank equals cell limit as x rightwards arrow 1 to the power of plus of x end cell row blank equals 1 end table

2) Limit kirinya adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 to the power of minus of g left parenthesis x right parenthesis end cell equals cell limit as x rightwards arrow 1 to the power of minus of square root of 1 minus x end root end cell row blank equals cell square root of 1 minus 1 end root end cell row blank equals cell square root of 0 end cell row blank equals 0 end table

Sehingga diperoleh limit kanannya 1 dan limit kirinya 0. Karena nilai limit kanan dan limit kirinya berbeda, maka fungsi tersebut tidak memiliki limit.

Dengan demikian, nilai limit dari limit as x rightwards arrow 1 of g left parenthesis x right parenthesis comma space untuk space g left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator x squared minus x over denominator x minus 1 end fraction end cell cell x less than 1 end cell row cell square root of 1 minus x end root end cell cell x greater or equal than 1 end cell end table close tidak ada.

0

Roboguru

Hitunglah setiap limit fungsi berikut. a.    b.

Pembahasan Soal:

a. limit as x rightwards arrow 2 of f left parenthesis x right parenthesis comma space untuk space f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell 3 minus 2 x end cell cell x less or equal than 2 end cell row cell x squared minus 5 end cell cell x greater than 2 end cell end table close

Fungsi f(x) terbagi menjadi dua interval, maka untuk menentukan x2limf(x), harus menentukan dulu limit kanan dan kiri sebagai berikut.

limx2f(x)====limx2(32x)32(2)341
 

limx2+f(x)====limx2+(x25)225451  

Berdasarkan uraian di atas, x2limf(x)=x2+limf(x)=1, maka x2limf(x)=1.

Dengan demikian, x2limf(x)=1.

b. limit as x rightwards arrow 1 of g left parenthesis x right parenthesis comma space untuk space g left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell fraction numerator x squared minus x over denominator x minus 1 end fraction end cell cell x less than 1 end cell row cell square root of 1 minus x end root end cell cell x greater or equal than 1 end cell end table close

Fungsi g(x) terbagi menjadi dua interval, maka untuk menentukan x1limg(x), harus menentukan dulu limit kanan dan kiri sebagai berikut.

limx1g(x)====limx1(x1x2x)limx1(x1)x(x1)limx1x1
 

limx1+g(x)====limx1+1x1100   

 

Berdasarkan uraian di atas, x1limg(x)=x1+limg(x), maka nilaix1limg(x) tidak ada.

Dengan demikian, nilaix1limg(x) tidak ada.

0

Roboguru

Tentukan nilai  yang memenuhi agar fungsi  berikut mempunyai limit. b.

Pembahasan Soal:

agar begin mathsize 14px style f open parentheses x close parentheses end style mempunyai nilai limit maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow a to the power of minus of f open parentheses x close parentheses end cell equals cell limit as x rightwards arrow a to the power of plus of f open parentheses x close parentheses space end cell row cell 1 half open parentheses a close parentheses plus 6 end cell equals cell 2 open parentheses a close parentheses plus 21 over 2 end cell row cell a over 2 plus 6 end cell equals cell 2 a plus 21 over 2 end cell row cell a over 2 minus 2 a end cell equals cell 21 over 2 minus 6 end cell row cell fraction numerator negative 3 a over denominator 2 end fraction end cell equals cell 9 over 2 end cell row cell negative 3 a end cell equals 9 row a equals cell negative 3 end cell end table end style 

Jadi nilai undefined yang memenuhi adalah -3

0

Roboguru

Diketahui fungsi . Pernyataan berikut yang benar adalah ....

Pembahasan Soal:

untuk begin mathsize 14px style x less than negative 2 comma space f open parentheses x close parentheses equals 3 x end style maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 to the power of minus of f open parentheses x close parentheses end cell equals cell 3 open parentheses negative 2 close parentheses end cell row blank equals cell negative 6 end cell end table end style

untuk begin mathsize 14px style x greater or equal than negative 2 comma space f left parenthesis x right parenthesis equals x plus 4 end style maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow negative 2 to the power of plus of f left parenthesis x right parenthesis end cell equals cell open parentheses negative 2 close parentheses plus 4 end cell row blank equals 2 end table end style

jadi begin mathsize 14px style limit as x rightwards arrow negative 2 of f open parentheses x close parentheses end style tidak ada karena limit kiri dan limit kanan di begin mathsize 14px style x equals negative 2 end style tidak sama 

Jadi tidak ada jawaban yang tepat

0

Roboguru

Selidiki fungsi tersebut mempunyai lmit atau tidak, berikan alasan! A).

Pembahasan Soal:

Diketahui begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator open vertical bar x minus 1 close vertical bar over denominator x minus 1 end fraction end style, maka:

- Untuk begin mathsize 14px style x rightwards arrow 1 to the power of plus end style 

begin mathsize 14px style limit as x rightwards arrow 1 to the power of plus of fraction numerator x minus 1 over denominator x minus 1 end fraction equals limit as x rightwards arrow 1 to the power of plus of space 1 equals 1 end style 

- Untuk begin mathsize 14px style x rightwards arrow 1 to the power of minus end style 

begin mathsize 14px style limit as x rightwards arrow 1 to the power of plus of fraction numerator negative open parentheses x minus 1 close parentheses over denominator x minus 1 end fraction equals limit as x rightwards arrow 1 to the power of plus of space open parentheses negative 1 close parentheses equals negative 1 end style 

sehingga:

begin mathsize 14px style limit as x rightwards arrow 1 to the power of plus of fraction numerator open vertical bar x minus 1 close vertical bar over denominator x minus 1 end fraction not equal to limit as x rightwards arrow 1 to the power of minus of fraction numerator open vertical bar x minus 1 close vertical bar over denominator x minus 1 end fraction end style 

Jadi, begin mathsize 14px style limit as x rightwards arrow 1 of fraction numerator open vertical bar x minus 1 close vertical bar over denominator x minus 1 end fraction end style tidak mempunyai limit karena nilai limit kanan begin mathsize 14px style not equal to end style limit kiri.

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