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Kerjakan limit berikut dengan metode subtitusi  !

Pertanyaan

Kerjakan limit berikut dengan metode subtitusi begin mathsize 14px style limit as x rightwards arrow 2 of fraction numerator 9 x plus 3 over denominator 2 x end fraction end style !

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator 9 x plus 3 over denominator 2 x end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator 9 times 2 plus 3 over denominator 2 times 2 end fraction end cell row blank equals cell fraction numerator 18 plus 3 over denominator 4 end fraction end cell row blank equals cell 21 over 4 end cell row blank equals cell 5 1 fourth end cell end table end style 

Jadi, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 2 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 9 x plus 3 over denominator 2 x end fraction 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 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 fourth end cell end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Damanhuri

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

Nilai dari  adalah ....

Pembahasan Soal:

Perhatikan bahwa jika dilakukan substitusi x equals 3 ke begin mathsize 14px style limit as straight x rightwards arrow 3 of fraction numerator straight x cubed minus 27 over denominator straight x minus 3 end fraction end style, maka didapatkan hasil undefined .

Oleh karena itu, perlu cara lain untuk mencari nilai limitnya.

Ingat bahwa a cubed minus b cubed equals open parentheses a minus b close parentheses open parentheses a squared plus a b plus b squared close parentheses.

Oleh karena itu, x cubed minus 27 dapat dituliskan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell x cubed minus 27 end cell equals cell x cubed minus 3 cubed end cell row blank equals cell open parentheses x minus 3 close parentheses open parentheses x squared plus x times 3 plus 3 squared close parentheses end cell row blank equals cell open parentheses x minus 3 close parentheses open parentheses x squared plus 3 x plus 9 close parentheses end cell end table 

Dengan demikian, nilai dari begin mathsize 14px style limit as straight x rightwards arrow 3 of fraction numerator straight x cubed minus 27 over denominator straight x minus 3 end fraction end style adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 3 of fraction numerator x cubed minus 27 over denominator x minus 3 end fraction end cell equals cell limit as x rightwards arrow 3 of fraction numerator open parentheses x minus 3 close parentheses open parentheses x squared plus 3 x plus 9 close parentheses over denominator x minus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 3 of fraction numerator up diagonal strike open parentheses x minus 3 close parentheses end strike open parentheses x squared plus 3 x plus 9 close parentheses over denominator up diagonal strike x minus 3 end strike end fraction end cell row blank equals cell limit as x rightwards arrow 3 of open parentheses x squared plus 3 x plus 9 close parentheses end cell row blank equals cell 3 squared plus 3 open parentheses 3 close parentheses plus 9 end cell row blank equals cell 9 plus 9 plus 9 end cell row blank equals 27 end table

Jadi, jawaban yang tepat adalah D.

Roboguru

Pembahasan Soal:

 Dengan aturan L'Hopital 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow c of fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction end cell equals cell limit as x rightwards arrow c of fraction numerator f apostrophe open parentheses x close parentheses over denominator g apostrophe open parentheses x close parentheses end fraction end cell row cell space space space space space space space space space space space space space space end cell equals cell limit as x rightwards arrow 1 of fraction numerator negative 1 over denominator 1 end fraction end cell row cell space space space space space space space space space space space space space space end cell equals cell negative 1 end cell end table end style

 

Roboguru

...

Pembahasan Soal:

Perhatikan bahwa jika dilakukan substitusi begin mathsize 14px style x equals 1 end style, maka didapatkan hasil begin mathsize 14px style 0 over 0 end style .
Sehingga perlu dicari cara lain untuk mencari nilai limitnya.

Misalkan begin mathsize 14px style straight x equals straight y to the power of 6 end style .
Maka untuk begin mathsize 14px style straight x rightwards arrow 1 end style   maka didapat bahwa begin mathsize 14px style straight y rightwards arrow 1 end style  .
Sehingga

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction end style 

Perhatikan bahwa begin mathsize 14px style square root of straight y to the power of 6 end root equals open vertical bar straight y cubed close vertical bar end style . Karena begin mathsize 14px style straight y rightwards arrow 1 comma space maka space straight y greater than 0 comma space sehingga square root of straight y to the power of 6 end root equals straight y cubed end style .Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 minus straight y cubed over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow 1 of fraction numerator open parentheses 1 minus straight y close parentheses open parentheses 1 plus straight y plus straight y squared close parentheses over denominator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y close parentheses end fraction equals limit as straight y rightwards arrow 1 of fraction numerator 1 plus straight y plus straight y squared over denominator 1 plus straight y end fraction equals fraction numerator 1 plus 1 plus 1 squared over denominator 1 plus 1 end fraction equals 3 over 2 end style 

Perhatikan bahwa untuk begin mathsize 14px style straight x rightwards arrow 1 end style , sebenarnya didapat pula bahwa begin mathsize 14px style straight y rightwards arrow 1 end style .
Sehingga

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction end style 

Perhatikan bahwa   begin mathsize 14px style square root of straight y to the power of 6 end root equals open vertical bar straight y cubed close vertical bar end style . Karena begin mathsize 14px style straight y rightwards arrow negative 1 comma space maka space straight y less than 0 comma space sehingga square root of straight y to the power of 6 end root equals negative straight y cubed end style .Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus square root of straight y to the power of 6 end root over denominator 1 minus cube root of straight y to the power of 6 end root end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus open parentheses negative straight y cubed close parentheses over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 plus straight y cubed over denominator 1 minus straight y squared end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y plus straight y squared close parentheses over denominator open parentheses 1 plus straight y close parentheses open parentheses 1 minus straight y close parentheses end fraction equals limit as straight y rightwards arrow negative 1 of fraction numerator 1 minus straight y plus straight y squared over denominator 1 minus straight y end fraction equals fraction numerator 1 minus open parentheses negative 1 close parentheses plus open parentheses negative 1 close parentheses squared over denominator 1 minus open parentheses negative 1 close parentheses end fraction equals fraction numerator 1 plus 1 plus 1 over denominator 1 plus 1 end fraction equals 3 over 2 end style 

Maka

begin mathsize 14px style limit as straight x rightwards arrow 1 of fraction numerator 1 minus square root of straight x over denominator 1 minus cube root of straight x end fraction equals 3 over 2 end style 

Jadi, jawabannya adalah D.

Roboguru

Pembahasan Soal:

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 2 of fraction numerator x squared minus 4 x plus 4 over denominator x minus 2 end fraction end cell equals cell limit as x rightwards arrow 2 of fraction numerator open parentheses x minus 2 close parentheses up diagonal strike open parentheses x minus 2 close parentheses end strike over denominator up diagonal strike open parentheses x minus 2 close parentheses end strike end fraction end cell row cell space space space space space space space space space space space space space space space space space space space space space space end cell equals cell open parentheses 2 minus 2 close parentheses end cell row cell space space space space space space space space space space space space space space space space space space space space space space end cell equals 0 end table end style 

 

 

Roboguru

....

Pembahasan Soal:

Perhatikan bahwa jika dilakukan substitusi = 5, maka didapatkan hasil undefined .
Sehingga perlu dicari cara lain untuk mencari nilai limitnya.

Perhatikan bahwa

begin mathsize 14px style limit as straight x rightwards arrow 5 of fraction numerator square root of 6 minus square root of straight x minus 1 end root end root minus square root of 2 plus square root of straight x minus 1 end root end root over denominator square root of straight x minus 1 end root minus 2 end fraction equals limit as straight x rightwards arrow 5 of open parentheses fraction numerator square root of 6 minus square root of straight x minus 1 end root end root minus square root of 2 plus square root of straight x minus 1 end root end root over denominator square root of straight x minus 1 end root minus 2 end fraction times fraction numerator square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root over denominator square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root end fraction close parentheses equals limit as straight x rightwards arrow 5 of fraction numerator open parentheses 6 minus square root of straight x minus 1 end root close parentheses minus open parentheses 2 plus square root of straight x minus 1 end root close parentheses over denominator open parentheses square root of straight x minus 1 end root minus 2 close parentheses open parentheses square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root close parentheses end fraction equals limit as straight x rightwards arrow 5 of fraction numerator 4 minus 2 square root of straight x plus 1 end root over denominator open parentheses square root of straight x minus 1 end root minus 2 close parentheses open parentheses square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root close parentheses end fraction equals limit as straight x rightwards arrow 5 of fraction numerator negative 2 open parentheses square root of straight x plus 1 end root minus 2 close parentheses over denominator open parentheses square root of straight x minus 1 end root minus 2 close parentheses open parentheses square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root close parentheses end fraction equals limit as straight x rightwards arrow 5 of fraction numerator negative 2 over denominator square root of 6 minus square root of straight x minus 1 end root end root plus square root of 2 plus square root of straight x minus 1 end root end root end fraction equals fraction numerator negative 2 over denominator square root of 6 minus square root of 5 minus 1 end root end root plus square root of 2 plus square root of 5 minus 1 end root end root end fraction equals fraction numerator negative 2 over denominator square root of 6 minus square root of 4 end root plus square root of 2 plus square root of 4 end root end fraction equals fraction numerator negative 2 over denominator square root of 6 minus 2 end root plus square root of 2 plus 2 end root end fraction equals fraction numerator negative 2 over denominator square root of 4 plus square root of 4 end fraction equals fraction numerator negative 2 over denominator 2 plus 2 end fraction equals fraction numerator negative 2 over denominator 4 end fraction equals negative 1 half end style 

Jadi, jawaban yang tepat adalah D.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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