Roboguru

Diketahui  dan . Tentukan :

Pertanyaan

Diketahui limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis equals 8 comma blank limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis equals negative 3 dan limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis equals 4. Tentukan :

limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight f open parentheses straight x close parentheses plus straight g open parentheses straight x close parentheses over denominator straight h open parentheses straight x close parentheses end fraction 

Pembahasan Soal:

Dalam menyelesaikan soal di atas, ingat sifat limit di bawah ini.

left parenthesis straight i right parenthesis space limit as x rightwards arrow c of invisible function application open square brackets f open parentheses x close parentheses plus-or-minus g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of invisible function application f left parenthesis x right parenthesis plus-or-minus limit as x rightwards arrow c of invisible function application g left parenthesis x right parenthesis left parenthesis ii right parenthesis space limit as x rightwards arrow c of invisible function application open square brackets fraction numerator f open parentheses x close parentheses over denominator g left parenthesis x right parenthesis end fraction close square brackets equals fraction numerator limit as x rightwards arrow c of invisible function application f left parenthesis x right parenthesis over denominator limit as x rightwards arrow c of invisible function application g left parenthesis x right parenthesis end fraction 


Pada soal, diketahui limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis equals 8 comma blank limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis equals negative 3, dan limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis equals 4. Maka penyelesaian soal di atas adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight f open parentheses straight x close parentheses plus straight g open parentheses straight x close parentheses over denominator straight h open parentheses straight x close parentheses end fraction end cell equals cell fraction numerator limit as straight x rightwards arrow straight a of invisible function application open square brackets straight f left parenthesis straight x right parenthesis plus straight g left parenthesis straight x right parenthesis close square brackets over denominator limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis end fraction end cell row blank equals cell fraction numerator limit as straight x rightwards arrow straight a of invisible function application straight f left parenthesis straight x right parenthesis plus limit as straight x rightwards arrow straight a of invisible function application straight g left parenthesis straight x right parenthesis over denominator limit as straight x rightwards arrow straight a of invisible function application straight h left parenthesis straight x right parenthesis end fraction end cell row blank equals cell fraction numerator 8 plus open parentheses negative 3 close parentheses over denominator 4 end fraction end cell row blank equals cell 5 over 4 end cell end table   


Jadi, nilai limit dari limit as straight x rightwards arrow straight a of invisible function application fraction numerator straight f open parentheses straight x close parentheses plus straight g open parentheses straight x close parentheses over denominator straight h open parentheses straight x close parentheses end fraction yang nilai limit masing-masing fungsinya diketahui adalah 5 over 4.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

Terakhir diupdate 01 Mei 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika , maka nilai

Pembahasan Soal:

Ingat Definisi turunan, jika

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell a x to the power of n rightwards arrow f apostrophe open parentheses x close parentheses equals a n x to the power of n minus 1 end exponent end cell end table

Penyelesaian dapat menggunakan limit fungsi dengan metode L'Hospital dan menggunakan turunan

Jika limit fungsi berbentuk limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals 0 over 0, maka limit fungsi tersebut bisa diselesaikan dengan menggunakan turunan, yaitu

limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction equals limit as x rightwards arrow k of fraction numerator f apostrophe open parentheses x close parentheses over denominator g apostrophe open parentheses x close parentheses end fraction

Menggunakan sifat limit berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow k of f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses end cell equals cell limit as x rightwards arrow k of f open parentheses x close parentheses plus-or-minus limit as x rightwards arrow k of g open parentheses x close parentheses end cell row cell limit as x rightwards arrow k of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction end cell equals cell fraction numerator limit as x rightwards arrow k of f open parentheses x close parentheses over denominator limit as x rightwards arrow k of g open parentheses x close parentheses end fraction end cell row cell limit as x rightwards arrow k of a f open parentheses x close parentheses end cell equals cell a limit as x rightwards arrow k of f open parentheses x close parentheses end cell end table

Serta memisalkan f open parentheses x close parentheses equals cube root of a x cubed plus b end root, didapatkan

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 half of open parentheses fraction numerator f open parentheses x close parentheses minus 2 over denominator x minus begin display style 1 half end style end fraction close parentheses end cell equals cell limit as x rightwards arrow 1 half of open parentheses fraction numerator f apostrophe open parentheses x close parentheses over denominator 1 end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell end table

Dengan demikian, dengan dalil L'Hospital diperoleh

 

table attributes columnalign right center left columnspacing 0px end attributes row blank blank blank row cell limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses end cell equals A end table

Akan dicari nilai dari limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses. Dengan menjabarkan bentuk aljabar sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction end cell equals cell fraction numerator cube root of begin display style 1 over 8 open parentheses a x cubed plus b close parentheses end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction end cell row blank equals cell fraction numerator cube root of begin display style 1 over 8 end style end root times cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x over denominator 4 x squared minus 1 end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x over denominator 4 x squared minus 1 end fraction end cell end table

Akan dicari nilai limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses dengan L'Hospital.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half cube root of open parentheses a x cubed plus b close parentheses end root minus 2 x end style over denominator 4 x squared minus 1 end fraction end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half f open parentheses x close parentheses minus 2 x end style over denominator 4 x squared minus 1 end fraction space end cell row blank equals cell limit as x rightwards arrow 1 half of fraction numerator begin display style 1 half f apostrophe open parentheses x close parentheses minus 2 end style over denominator 8 x end fraction space end cell row blank equals cell fraction numerator begin display style limit as x rightwards arrow 1 half of 1 half f apostrophe open parentheses x close parentheses minus limit as x rightwards arrow 1 half of 2 end style over denominator limit as x rightwards arrow 1 half of 8 x end fraction end cell row blank equals cell fraction numerator begin display style 1 half limit as x rightwards arrow 1 half of f apostrophe open parentheses x close parentheses minus 2 end style over denominator 8 limit as x rightwards arrow 1 half of x end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style A minus begin display style 2 end style over denominator 8 times begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator begin display style 1 half open parentheses A minus 4 close parentheses end style over denominator 4 end fraction end cell row blank equals cell fraction numerator begin display style A minus 4 end style over denominator 8 end fraction end cell end table

Jadi, nilai  darilimit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses equals fraction numerator A minus 4 over denominator 8 end fraction.

Dengan demikian, ka limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of a x cubed plus b end root minus 2 over denominator x minus begin display style 1 half end style end fraction close parentheses equals A, maka nilai limit as x rightwards arrow 1 half of open parentheses fraction numerator cube root of begin display style fraction numerator a x cubed over denominator 8 end fraction plus b over 8 end style end root minus 2 x over denominator 4 x squared minus begin display style 1 end style end fraction close parentheses equals fraction numerator A minus 4 over denominator 8 end fraction

Roboguru

Pembahasan Soal:

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.

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.

Roboguru

Jika , nilai dari  adalah....

Pembahasan Soal:

Ingat bahwa:

limxa(q(x)p(x))limxap(x)+q(x)==limxaq(x)limxap(x)limxap(x)+limxaq(x)

Diketahui xalimg(x)= sehingga diperoleh

limxa[g(x)1+3]===limxag(x)1+limxa3limxag(x)limxa1+31+3

Jika penyebut bernilai tak hingga, maka nilai pecahannya akan mendekati 0, maka diperoleh

1+3==0+33

Dengan demikian, nilai dari xalim[g(x)1+3]=3.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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