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Jika , nilai  adalah ...

Pertanyaan

Jika size 14px lim with size 14px z size 14px rightwards arrow size 14px 1 below fraction numerator size 14px z to the power of size 14px 2 size 14px plus size 14px z size 14px minus size 14px 3 over denominator size 14px z size 14px plus size 14px 1 end fraction size 14px equals size 14px a size 14px minus size 14px 1, nilai begin mathsize 14px style a end style adalah ...        

  1. size 14px minus size 14px 1 

  2. size 14px minus size 14px 1 over size 14px 2   

  3. size 14px 0   

  4. size 14px 1 over size 14px 2  

  5. size 14px 1   

Pembahasan Soal:

Substitusikan nilai pendekatan z, 

limz1z+1z2+z3====1+112+1321+132121

Diketahui bahwa: size 14px lim with size 14px z size 14px rightwards arrow size 14px 1 below fraction numerator size 14px z to the power of size 14px 2 size 14px plus size 14px z size 14px minus size 14px 3 over denominator size 14px z size 14px plus size 14px 1 end fraction size 14px equals size 14px a size 14px minus size 14px 1, maka:

21aaa====a121+121+221

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Bella

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Nilai  adalah ...

Pembahasan Soal:

Limit merupakan sebuah konsep matematika dimana sesuatu dikatakan "hampir" atau "mendekati" nilai suatu bilangan tertentu.

 Dengan menggunakan dalil L'Hopital, dimana:

begin mathsize 14px style limit as x rightwards arrow a of fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow a of fraction numerator f apostrophe left parenthesis x right parenthesis over denominator g apostrophe left parenthesis x right parenthesis end fraction end style

maka tentukan turunan dari begin mathsize 14px style 2 x squared minus 162 end style dan begin mathsize 14px style x plus 3 square root of x minus 18 end style.

Turunan fungsi begin mathsize 14px style 2 x squared minus 162 end style:

begin mathsize 14px style y equals 2 x squared minus 162 y apostrophe equals 4 x end style 

Turunan fungsi begin mathsize 14px style x plus 3 square root of x minus 18 end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell x plus 3 square root of x minus 18 end cell row y equals cell x plus 3 x to the power of begin inline style 1 half end style end exponent minus 18 end cell row cell y apostrophe end cell equals cell 1 plus 3 times 1 half x to the power of begin inline style 1 half end style minus 1 end exponent end cell row cell y apostrophe end cell equals cell 1 plus 3 over 2 x to the power of negative begin inline style 1 half end style end exponent end cell row cell y apostrophe end cell equals cell 1 plus 3 fraction numerator 3 over denominator 2 square root of x end fraction end cell end table end style  

Nilai begin mathsize 14px style limit as x rightwards arrow 9 of space fraction numerator 2 x squared minus 162 over denominator x plus 3 square root of x minus 18 end fraction end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 9 of space fraction numerator 2 x squared minus 162 over denominator x plus 3 square root of x minus 18 end fraction end cell equals cell limit as x rightwards arrow 9 of space fraction numerator 4 x over denominator 1 plus begin display style fraction numerator 3 over denominator 2 square root of x end fraction end style end fraction end cell row blank equals cell fraction numerator 4 left parenthesis 9 right parenthesis over denominator 1 plus begin display style fraction numerator 3 over denominator 2 square root of 9 end fraction end style end fraction end cell row blank equals cell fraction numerator 36 over denominator 1 plus begin display style fraction numerator 3 over denominator 2 times 3 end fraction end style end fraction end cell row blank equals cell fraction numerator 36 over denominator 1 plus begin display style 3 over 6 end style end fraction end cell row blank equals cell fraction numerator 36 over denominator begin display style 9 over 6 end style end fraction end cell row blank equals cell fraction numerator 36 over denominator begin display style 3 over 2 end style end fraction end cell row blank equals cell 36 cross times 2 over 3 end cell row blank equals 24 end table end style  

Sehingga, nilai dari begin mathsize 14px style limit as x rightwards arrow 9 of space fraction numerator 2 x squared minus 162 over denominator x plus 3 square root of x minus 18 end fraction end style adalah begin mathsize 14px style 24 end style

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Nilai  adalah ....

Pembahasan Soal:

Limit merupakan sebuah konsep matematika dimana sesuatu dikatakan "hampir" atau "mendekati" nilai suatu bilangan tertentu.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of space fraction numerator 6 x square root of x minus 24 square root of x over denominator square root of x minus 2 end fraction end cell equals cell limit as x rightwards arrow 4 of space fraction numerator 6 square root of x open parentheses x minus 4 close parentheses over denominator square root of x minus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 6 square root of x left parenthesis x minus 4 right parenthesis over denominator square root of x minus 2 end fraction cross times fraction numerator square root of x plus 2 over denominator square root of x plus 2 end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space fraction numerator 6 square root of x up diagonal strike left parenthesis x minus 4 right parenthesis end strike open parentheses square root of x plus 2 close parentheses over denominator up diagonal strike x minus 4 end strike end fraction end cell row blank equals cell limit as x rightwards arrow 4 of space 6 square root of x open parentheses square root of x plus 2 close parentheses end cell row blank equals cell 6 square root of 4 open parentheses square root of 4 plus 2 close parentheses end cell row blank equals cell 6 times 2 open parentheses 2 plus 2 close parentheses end cell row blank equals cell 12 times 4 end cell row blank equals 48 end table end style 

Jadi, nilai dari begin mathsize 14px style limit as x rightwards arrow 4 of space fraction numerator 6 x square root of x minus 24 square root of x over denominator square root of x minus 2 end fraction end style adalah begin mathsize 14px style 48 end style

1

Roboguru

...

Pembahasan Soal:

Gunakan pemfaktoran untuk menyederhanakan fungsi limit.

begin mathsize 14px style table attributes columnalign center center left end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x squared plus 5 x minus 14 over denominator x squared plus 11 x plus 28 end fraction end cell equals cell limit as x rightwards arrow negative 7 of space fraction numerator x squared plus 7 x minus 2 x minus 14 over denominator x squared plus 7 x plus 4 x plus 28 end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator x left parenthesis x plus 7 right parenthesis minus 2 left parenthesis x plus 7 right parenthesis over denominator x left parenthesis x plus 7 right parenthesis plus 4 left parenthesis x plus 7 right parenthesis end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator down diagonal strike left parenthesis x plus 7 right parenthesis end strike left parenthesis x minus 2 right parenthesis over denominator down diagonal strike left parenthesis x plus 7 right parenthesis end strike left parenthesis x plus 4 right parenthesis end fraction end cell row space equals cell limit as x rightwards arrow negative 7 of space fraction numerator x minus 2 over denominator x plus 4 end fraction end cell row space space space row space space space row space space cell space space end cell end table end style 

Kemudian substitusi begin mathsize 14px style x equals negative 7 end style ke seperti berikut:

begin mathsize 14px style table attributes columnalign center center left end attributes row cell limit as x rightwards arrow negative 7 of space fraction numerator x minus 2 over denominator x plus 4 end fraction end cell equals cell fraction numerator open parentheses negative 7 close parentheses minus 2 over denominator open parentheses negative 7 close parentheses plus 4 end fraction end cell row space equals cell fraction numerator negative 9 over denominator negative 3 end fraction end cell row space equals 3 end table end style 

 

0

Roboguru

Untuk x mendekati 1, nilai limit   dengan  secara intuitif adalah ...

Pembahasan Soal:

1

Roboguru

Nilai

Pembahasan Soal:

Limit dapat diartikan sebagai menuju suatu batas, sesuatu yang dekat namun tidak dapat dicapai. Cara penyelesaian nilai undefined mendekati berhingga adalah dengan substitusi, pemfaktoran, dan dikalikan dengan sekawannya.

Persoalan di atas dapat digunakan metode substitusi.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of fraction numerator 3 x plus 4 x to the power of negative 1 end exponent over denominator 4 x minus x to the power of negative 1 end exponent end fraction end cell equals cell limit as x rightwards arrow 0 of fraction numerator 3 x squared plus 4 over denominator 4 x squared minus 1 end fraction end cell row blank equals cell fraction numerator 3 open parentheses 0 close parentheses squared plus 4 over denominator 4 open parentheses 0 close parentheses squared minus 1 end fraction end cell row blank equals cell fraction numerator 4 over denominator negative 1 end fraction end cell row blank equals cell negative 4 end cell end table end style 

Jadi, diperoleh begin mathsize 14px style limit as x rightwards arrow 0 of fraction numerator 3 x plus 4 x to the power of negative 1 end exponent over denominator 4 x minus x to the power of negative 1 end exponent end fraction equals negative 4 end style.

0

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