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Tentukan nilai limit berikut. a. x→5lim​(−2x+7)7

Pertanyaan

Tentukan nilai limit berikut.

a. x5lim(2x+7)7 

Pembahasan Soal:

untuk menentukan nilai limitnya, substitusikan nilai x=5, sehingga:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 5 of left parenthesis negative 2 x plus 7 right parenthesis to the power of 7 end cell equals cell open parentheses negative 2 open parentheses 5 close parentheses plus 7 close parentheses to the power of 7 space end cell row blank equals cell open parentheses negative 10 plus 7 close parentheses to the power of 7 end cell row blank equals cell open parentheses negative 3 close parentheses to the power of 7 end cell row blank equals cell negative 2187 end cell end table end style  

Dengan demikian, nilai dari x5lim(2x+7)7 adalah 2187.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Indah

Mahasiswa/Alumni Universitas Lampung

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Nilai  adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell limit as x rightwards arrow c of left parenthesis x minus 3 right parenthesis squared end cell equals cell open parentheses limit as x rightwards arrow c of left parenthesis x minus 3 right parenthesis close parentheses squared end cell row blank equals cell open parentheses c minus 3 close parentheses squared end cell row blank equals cell c squared minus 6 c plus 9 end cell end table end style

Jadi nilai begin mathsize 14px style limit as x rightwards arrow c of left parenthesis x minus 3 right parenthesis squared equals c to the power of 2 end exponent minus 6 c plus 9 end style 

1

Roboguru

Diketahui  dan nilai limit berikut.   Nilai limit yang benar adalah ...

Pembahasan Soal:

Ingatlah sifat-sifat limit fungsi untuk menjawab soal di atas.

Akan dicoba satu-satu pernyataan pada soal apakah benar atau tidak. Diketahui limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis equals 3 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis straight i right parenthesis space space limit as straight x rightwards arrow 2 of invisible function application open square brackets straight f left parenthesis straight x right parenthesis minus straight f squared left parenthesis straight x right parenthesis close square brackets end cell equals cell limit as straight x rightwards arrow 2 of invisible function application f left parenthesis x right parenthesis minus open square brackets limit as straight x rightwards arrow 2 of invisible function application minus straight f left parenthesis straight x right parenthesis close square brackets squared end cell row blank equals cell 3 minus open parentheses 3 close parentheses squared end cell row blank equals cell 3 minus 9 end cell row blank equals cell negative 6. end cell end table

Pernyataan (i) salah.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis ii right parenthesis space space limit as straight x rightwards arrow 2 of invisible function application open square brackets open parentheses straight f left parenthesis straight x right parenthesis plus 2 close parentheses squared minus 4 space straight f left parenthesis straight x right parenthesis close square brackets end cell equals cell open square brackets space limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis plus 2 close square brackets squared minus 4 limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis end cell row blank equals cell open square brackets space limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis plus space limit as straight x rightwards arrow 2 of invisible function application 2 close square brackets squared minus 4 limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis end cell row blank equals cell open square brackets 3 plus 2 close square brackets squared minus 4 open parentheses 3 close parentheses end cell row blank equals cell 25 minus 12 end cell row blank equals cell 13. end cell end table 

Pernyataan (ii) benar.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis iii right parenthesis space space limit as straight x rightwards arrow 2 of invisible function application open square brackets straight f left parenthesis straight x right parenthesis minus 2 straight x plus 4 close square brackets squared end cell equals cell space limit as straight x rightwards arrow 2 of invisible function application open square brackets straight f left parenthesis straight x right parenthesis minus 2 straight x plus 4 close square brackets squared end cell row blank equals cell open square brackets limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis minus 2 straight x plus 4 close square brackets squared end cell row blank equals cell open square brackets limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis minus 2 limit as straight x rightwards arrow 2 of invisible function application straight x plus limit as straight x rightwards arrow 2 of invisible function application 4 close square brackets squared end cell row blank equals cell open square brackets 3 minus 2 open parentheses 2 close parentheses plus 4 close square brackets squared end cell row blank equals cell open square brackets 3 minus 4 plus 4 close square brackets squared end cell row blank equals cell 9. end cell end table 

Pertanyaan (iii) salah.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis iv right parenthesis space space limit as straight x rightwards arrow 2 of invisible function application cube root of 4 straight f squared left parenthesis straight x right parenthesis minus 9 end root end cell equals cell open square brackets limit as straight x rightwards arrow 2 of invisible function application 4 straight f squared left parenthesis straight x right parenthesis minus 9 close square brackets to the power of 1 third end exponent end cell row blank equals cell open square brackets 4 limit as straight x rightwards arrow 2 of invisible function application straight f squared left parenthesis straight x right parenthesis minus limit as straight x rightwards arrow 2 of invisible function application 9 close square brackets to the power of 1 third end exponent end cell row blank equals cell open square brackets 4 open parentheses limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis close parentheses squared minus limit as straight x rightwards arrow 2 of invisible function application 9 close square brackets to the power of 1 third end exponent end cell row blank equals cell open square brackets 4 open parentheses 3 close parentheses squared minus 9 close square brackets to the power of 1 third end exponent end cell row blank equals cell cube root of 36 minus 9 end root end cell row blank equals cell cube root of 27 end cell row blank equals cell 3. end cell end table   

Pertanyaan (iv) benar.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis straight v right parenthesis space space space limit as straight x rightwards arrow 2 of invisible function application fraction numerator straight f left parenthesis straight x right parenthesis plus 9 over denominator 1 minus straight f left parenthesis straight x right parenthesis end fraction end cell equals cell fraction numerator limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis plus 9 over denominator limit as straight x rightwards arrow 2 of invisible function application 1 minus straight f left parenthesis straight x right parenthesis end fraction end cell row blank equals cell fraction numerator limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis plus limit as straight x rightwards arrow 2 of invisible function application 9 over denominator limit as straight x rightwards arrow 2 of invisible function application 1 minus limit as straight x rightwards arrow 2 of invisible function application straight f left parenthesis straight x right parenthesis end fraction end cell row blank equals cell fraction numerator 3 plus 9 over denominator 1 minus 3 end fraction end cell row blank equals cell fraction numerator 12 over denominator negative 2 end fraction end cell row blank equals cell negative 6. end cell end table 

Pernyataan (v) benar.

Berdasarkan uraian di atas, maka pernyataan yang benar adalah (ii), (iv), dan (v).

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Jika L, K adalah bilangan real dan ,  maka tentukan .

Pembahasan Soal:

Ingat sifat limit:

  1. limit as x rightwards arrow a of c equals c
  2. limit as x rightwards arrow a of open parentheses f left parenthesis x right parenthesis plus-or-minus g left parenthesis x right parenthesis close parentheses equals limit as x rightwards arrow a of f left parenthesis x right parenthesis plus-or-minus limit as x rightwards arrow a of g left parenthesis x right parenthesis
  3. limit as x rightwards arrow a of open parentheses fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction close parentheses equals fraction numerator limit as x rightwards arrow a of f left parenthesis x right parenthesis over denominator limit as x rightwards arrow a of g left parenthesis x right parenthesis end fraction
  4. limit as x rightwards arrow a of open parentheses f left parenthesis x right parenthesis close parentheses to the power of straight n equals open parentheses limit as x rightwards arrow a of f left parenthesis x right parenthesis close parentheses to the power of n

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator f squared left parenthesis x right parenthesis minus straight L squared over denominator f squared left parenthesis x right parenthesis plus straight L squared end fraction end cell equals cell fraction numerator open parentheses limit as x rightwards arrow 2 of f left parenthesis x right parenthesis close parentheses squared minus limit as x rightwards arrow 2 of straight L squared over denominator open parentheses limit as x rightwards arrow 2 of f left parenthesis x right parenthesis close parentheses squared plus limit as x rightwards arrow 2 of straight L squared end fraction end cell row blank equals cell fraction numerator straight L minus straight L squared over denominator straight L plus straight L squared end fraction end cell row blank equals cell fraction numerator straight L left parenthesis 1 minus straight L right parenthesis over denominator straight L left parenthesis 1 plus straight L right parenthesis end fraction end cell end table


Jadi nilai table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator f squared left parenthesis x right parenthesis minus straight L squared over denominator f squared left parenthesis x right parenthesis plus straight L squared end fraction end cell equals cell fraction numerator straight L left parenthesis 1 minus straight L right parenthesis over denominator straight L left parenthesis 1 plus straight L right parenthesis end fraction end cell end table.

0

Roboguru

Dengan teorema limit, hitunglah :

Pembahasan Soal:

Ingat kembali sifat limit berikut.

  1. begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses fraction numerator straight f open parentheses straight x close parentheses over denominator straight g open parentheses straight x close parentheses end fraction close parentheses equals fraction numerator limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses close parentheses over denominator limit as straight x rightwards arrow straight c of open parentheses straight g open parentheses straight x close parentheses close parentheses end fraction end style 
  2. begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses cross times straight g open parentheses straight x close parentheses close parentheses equals limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses close parentheses cross times limit as straight x rightwards arrow straight c of open parentheses straight g open parentheses straight x close parentheses close parentheses end style 
  3. begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses straight f open parentheses straight x close parentheses plus-or-minus straight g open parentheses straight x close parentheses close parentheses equals limit as straight x rightwards arrow straight c of straight f open parentheses straight x close parentheses plus-or-minus limit as straight x rightwards arrow straight c of straight g open parentheses straight x close parentheses end style
  4. begin mathsize 14px style limit as straight x rightwards arrow straight c of straight k times straight f open parentheses straight x close parentheses equals straight k times limit as straight x rightwards arrow straight c of straight f open parentheses straight x close parentheses end style
  5. begin mathsize 14px style limit as straight x rightwards arrow straight c of open parentheses straight k close parentheses equals straight k end style  

Maka dengan menggunakan konsep sifat-sifat limit kemudian dengan mensubtitusi nilai straight x ke dalam limit fungsi tersebut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as straight x rightwards arrow 3 of fraction numerator open parentheses straight x minus 5 close parentheses cubed open parentheses 2 straight x minus 9 close parentheses over denominator straight x squared plus 2 straight x minus 3 end fraction end cell row blank equals cell fraction numerator open parentheses limit as straight x rightwards arrow 3 of open parentheses straight x close parentheses minus limit as straight x rightwards arrow 3 of open parentheses 5 close parentheses close parentheses cubed cross times open parentheses 2 cross times limit as straight x rightwards arrow 3 of open parentheses straight x close parentheses minus limit as straight x rightwards arrow 3 of open parentheses 9 close parentheses close parentheses over denominator open parentheses limit as straight x rightwards arrow 3 of open parentheses straight x close parentheses close parentheses squared plus open parentheses 2 cross times limit as straight x rightwards arrow 3 of open parentheses straight x close parentheses close parentheses minus limit as straight x rightwards arrow 3 of open parentheses 3 close parentheses end fraction end cell row blank equals cell fraction numerator open parentheses 3 minus 5 close parentheses cubed cross times open parentheses open parentheses 2 straight x 3 close parentheses minus 9 close parentheses over denominator 3 squared plus open parentheses 2 cross times 3 close parentheses minus 3 end fraction end cell row blank equals cell fraction numerator open parentheses negative 8 close parentheses cross times open parentheses negative 3 close parentheses over denominator 12 end fraction end cell row blank equals 2 end table 
 

Jadi, hasil dari begin mathsize 14px style limit as straight x rightwards arrow 3 of fraction numerator open parentheses straight x minus 5 close parentheses cubed open parentheses 2 straight x minus 9 close parentheses over denominator straight x squared plus 2 straight x minus 3 end fraction end style adalah 2.

0

Roboguru

Nilai  ....

Pembahasan Soal:

Perhatikan 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space open parentheses x squared plus 9 close parentheses to the power of 1 half end exponent end cell equals cell open parentheses 4 squared plus 9 close parentheses to the power of 1 half end exponent end cell row blank equals cell 25 to the power of 1 half end exponent end cell row blank equals 5 end table end style

0

Roboguru

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