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Pertanyaan

begin mathsize 14px style limit as x rightwards arrow 1 of open parentheses square root of 5 x plus 4 end root minus square root of 16 minus 7 x end root close parentheses equals space... space end style         

  1. undefined 

  2. begin mathsize 14px style negative 3 end style 

  3. begin mathsize 14px style 0 end style 

  4. begin mathsize 14px style 3 end style 

  5. undefined 

Pembahasan Soal:

limx1(5x+4167x)====51+416715+4167990 

Oleh karena itu, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Mahasiswa/Alumni Universitas Siliwangi

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Jika . Carilah setiap limit berikut. a.

Pembahasan Soal:

Sifat-sifat limit :

1. begin mathsize 14px style limit as x rightwards arrow c of open parentheses f open parentheses x close parentheses plus-or-minus g open parentheses x close parentheses close parentheses equals limit as x rightwards arrow c of f open parentheses x close parentheses plus-or-minus limit as x rightwards arrow c of g open parentheses x close parentheses end style

2. begin mathsize 14px style limit as x rightwards arrow c of f to the power of n open parentheses x close parentheses equals open parentheses limit as x rightwards arrow c of f open parentheses x close parentheses close parentheses to the power of n end style

3. begin mathsize 14px style limit as x rightwards arrow c of n-th root of f open parentheses x close parentheses end root equals n-th root of limit as x rightwards arrow c of f open parentheses x close parentheses end root end style , asalkan begin mathsize 14px style limit as x rightwards arrow c of f open parentheses x close parentheses greater or equal than 0 comma space straight n space genap end style 

Dengan menggunakan sifat-sifat limit fungsi diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow c of square root of f squared left parenthesis x right parenthesis minus g squared left parenthesis x right parenthesis end root end cell equals cell square root of limit as x rightwards arrow c of space open parentheses f squared left parenthesis x right parenthesis minus g squared left parenthesis x right parenthesis close parentheses end root end cell row blank equals cell square root of limit as x rightwards arrow c of space f squared left parenthesis x right parenthesis minus limit as x rightwards arrow c of space g squared left parenthesis x right parenthesis end root end cell row blank equals cell square root of open parentheses limit as x rightwards arrow c of space f left parenthesis x right parenthesis close parentheses squared minus open parentheses limit as x rightwards arrow c of space g left parenthesis x right parenthesis close parentheses squared end root end cell row blank equals cell square root of 6 squared minus left parenthesis negative 8 right parenthesis squared end root end cell row blank equals cell square root of 36 minus 64 end root end cell row blank equals cell square root of negative 28 end root end cell row blank equals cell 2 square root of negative 7 end root end cell end table end style 

Dengan demikian, diperoleh begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow c of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of f squared left parenthesis x right parenthesis minus g squared left parenthesis x right parenthesis end root end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of negative 7 end root end cell end table end style.

0

Roboguru

Diberikan x→alim​f(x)=3 dan .  Hitunglah nilai setiap limit berikut. a.   b.   c.   d.

Pembahasan Soal:

Ingat bahwa:

Limit memiliki sifat seperti di bawah ini.

1. xalim[f(x)+g(x)]=xalimf(x)+xalimg(x)

2. xalim[f(x)g(x)]=xalimf(x)xalimg(x)

3. xalim[f(x)×g(x)]=xalimf(x)×xalimg(x)

4. xalimkf(x)=k×xalimf(x)

5. xalim[g(x)f(x)]=limxag(x)limxaf(x)

6. xalim[f(x)]n=[xalimf(x)]n

7. xalimf(x)=xalimf(x)

8. xalimc=c,cR


Diberikan xalimf(x)=3danxalimg(x)=1

a. xalimf2(x)+g2(x)

Berdasarkan sifat 1, 6, dan 7 diperoleh:

limxaf2(x)+g2(x)====[limxaf(x)]2+[limxag(x)]2[3]2+[1]29+110

Dengan demikian, hasil dari xalimf2(x)+g2(x) adalah 10.

b. xalimf(x)+g(x)2f(x)3g(x)

Berdasarkan sifat 1, 2, 4, dan 5 diperoleh:

limxaf(x)+g(x)2f(x)3g(x)====limxaf(x)+limxag(x)2limxaf(x)3limxag(x)3+(1)2[3]3[1]26+329

Dengan demikian, hasil dari xalimf(x)+g(x)2f(x)3g(x) adalah 29.

c. xalim3g(x)[f(x)+3]

Berdasarkan sifat 1, 3, 7, dan 8 diperoleh:

limxa3g(x)[f(x)+3]====3limxag(x)×[limxaf(x)+limxa3]31×[3+3]1×66

Dengan demikian, hasil dari xalim3g(x)[f(x)+3] adalah 6.

d. xalim[f(x)3]4

Berdasarkan sifat 1, 6, dan 8 diperoleh:

limxa[f(x)3]4=====[limxa[f(x)3]]4[limxaf(x)limxa3]4[33]4[0]40

Dengan demikian, hasil dari xalim[f(x)3]4 adalah 0.

0

Roboguru

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

0

Roboguru

Diketahui , , dan . Tentukan .

Pembahasan Soal:

Ingat sifat limit bahwa 

begin mathsize 14px style limit as x rightwards arrow a of left parenthesis f left parenthesis x right parenthesis plus-or-minus g left parenthesis x right parenthesis right parenthesis 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 end style,

maka

begin mathsize 14px style limit as x rightwards arrow a of left parenthesis 2 f left parenthesis x right parenthesis plus 3 g left parenthesis x right parenthesis minus square root of h left parenthesis x right parenthesis end root right parenthesis equals limit as x rightwards arrow a of 2 f left parenthesis x right parenthesis plus limit as x rightwards arrow a of 3 g left parenthesis x right parenthesis minus limit as x rightwards arrow a of square root of h left parenthesis x right parenthesis end root equals stack 2 times lim space f left parenthesis x right parenthesis with x rightwards arrow a below plus stack 3 times lim with x rightwards arrow a below g left parenthesis x right parenthesis minus square root of limit as x rightwards arrow a of h left parenthesis x right parenthesis end root equals 2 times 8 plus 3 times left parenthesis negative 3 right parenthesis minus square root of 4 equals 16 plus negative 9 minus 2 equals 5 end style

Jadi hasil dari begin mathsize 14px style limit as x rightwards arrow a of left parenthesis 2 f left parenthesis x right parenthesis plus 3 g left parenthesis x right parenthesis minus square root of h left parenthesis x right parenthesis end root right parenthesis end style adalah begin mathsize 14px style 5 end style.

0

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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