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Nilai dari  adalah ....

Pertanyaan

Nilai dari limit as x rightwards arrow 0 of space fraction numerator 3 x to the power of 5 minus 3 x cubed plus 2 x over denominator 2 x minus 2 x squared minus 2 x to the power of 5 end fraction adalah ....

  1. negative 2 

  2. 0 

  3. 1 

  4. 3 over 2 

  5. 2 

Pembahasan Soal:

Limit fungsi polinomial di atas dapat dihitung seperti berikut,

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator 3 x to the power of 5 minus 3 x cubed plus 2 x over denominator 2 x minus 2 x squared minus 2 x to the power of 5 end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator up diagonal strike x open parentheses 3 x to the power of 4 minus 3 x squared plus 2 close parentheses over denominator 2 up diagonal strike x open parentheses 1 minus x minus x to the power of 4 close parentheses end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses 3 x to the power of 4 minus 3 x squared plus 2 close parentheses over denominator 2 open parentheses 1 minus x minus x to the power of 4 close parentheses end fraction end cell row blank equals cell fraction numerator 3 times 0 to the power of 4 minus 3 times 0 squared plus 2 over denominator 2 open parentheses 1 minus 0 minus 0 to the power of 4 close parentheses end fraction end cell row blank equals cell fraction numerator 2 over denominator 2 left parenthesis 1 right parenthesis end fraction end cell row blank equals 1 end table  

Dengan demikian, nilai dari limit di atas adalah 1.

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Kumaralalita

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 01 Mei 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.

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

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

...

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

Pembahasan Soal:

Limit

table attributes columnalign right center left columnspacing 0px end attributes row cell lim with italic x yields minus sign 1 below fraction numerator italic x squared minus sign 1 over denominator italic x plus 1 end fraction end cell equals cell lim with italic x yields minus sign 1 below fraction numerator open parentheses italic x minus sign 1 close parentheses open parentheses italic x plus 1 close parentheses over denominator italic x plus 1 end fraction end cell row blank equals cell lim with italic x yields minus sign 1 below open parentheses italic x minus sign 1 close parentheses end cell row blank equals cell open parentheses negative sign 1 minus sign 1 close parentheses end cell row blank equals cell negative sign 2 end cell end table

Jadi, limit as x rightwards arrow negative 1 of fraction numerator x squared minus 1 over denominator x plus 1 end fractionequals minus sign 2

0

Roboguru

Find the value of the limit and which applicable limit theorems used to find the value.  d.

Pembahasan Soal:

0

Roboguru

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