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Tentukan nilai a yang memenuhi kesamaan berikut!

Pertanyaan

Tentukan nilai a yang memenuhi kesamaan berikut!

limit as x rightwards arrow straight a of invisible function application fraction numerator 2 x minus 6 over denominator 8 minus x end fraction equals 4 over 3

Pembahasan Soal:

Gunakanlah metode substitusi untuk menyelesaikan limit tersebut, lalu jalankan operasi hitung hingga menghasilkan nilai a.

 table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow straight a of invisible function application fraction numerator 2 x minus 6 over denominator 8 minus x end fraction end cell equals cell 4 over 3 end cell row cell fraction numerator 2 open parentheses straight a close parentheses minus 6 over denominator 8 minus straight a end fraction end cell equals cell 4 over 3 end cell row cell fraction numerator 2 straight a minus 6 over denominator 8 minus straight a end fraction end cell equals cell 4 over 3 end cell row cell 4 open parentheses 8 minus straight a close parentheses end cell equals cell 3 open parentheses 2 straight a minus 6 close parentheses end cell row cell 32 minus 4 straight a end cell equals cell 6 straight a minus 18 end cell row cell 6 straight a plus 4 straight a end cell equals cell 32 plus 18 end cell row cell 10 straight a end cell equals 50 row straight a equals cell 50 over 10 end cell row straight a equals 5 end table 
 

Jadi, nilai a yang memenuhi persamaan limit as x rightwards arrow straight a of invisible function application fraction numerator 2 x minus 6 over denominator 8 minus x end fraction equals 4 over 3 adalah 5.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

Terakhir diupdate 01 Mei 2021

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Pertanyaan yang serupa

Pembahasan Soal:

kita coba substitusi begin mathsize 14px style x equals 2 end style, apabila kita mendapatkan hasil yang bukan merupakan bentuk tak tentu. Maka, itulah nilai limitnya.

begin mathsize 14px style space space limit as x rightwards arrow 2 of square root of 2 x squared minus 3 x minus 2 end root equals square root of 2.2 squared minus 3.2 minus 2 end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 8 minus 6 minus 2 end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 0 end style 

Jadi, nilai limitnya adalah 0.

Roboguru

Tentukan nilai .

Pembahasan Soal:

Berdasarkan sifat limit yaitu:

limit as x rightwards arrow k of space fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals fraction numerator limit as x rightwards arrow k of space f left parenthesis x right parenthesis over denominator limit as x rightwards arrow k of g left parenthesis x right parenthesis end fraction comma space g left parenthesis x right parenthesis not equal to 0 

Dengan menggunakan cara substitusi, substitusikan nilai x equals 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 2 of fraction numerator straight x squared plus 2 ax space plus straight b over denominator straight x minus 2 end fraction end cell equals cell negative 3 end cell row cell fraction numerator open parentheses 2 close parentheses squared plus 2 a open parentheses 2 close parentheses space plus straight b over denominator 2 minus 2 end fraction end cell equals cell negative 3 end cell row cell fraction numerator 4 plus 4 a space plus straight b over denominator 0 end fraction end cell equals cell negative 3 end cell end table

Karena dengan mensubstitusikan x equals 2 limit tersebut tidak terdefinisi.maka menggunakan cara lain untuk menyelesaikan permasalahan di atas.

begin mathsize 14px style limit as straight x rightwards arrow 2 of fraction numerator straight x squared plus 2 ax space plus straight b over denominator straight x minus 2 end fraction end style mempunyai nilai jika Error converting from MathML to accessible text. untuk begin mathsize 14px style x equals 2 end style, Maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 squared plus 2 straight a times 2 space plus straight b end cell equals 0 row cell 4 plus 4 straight a plus straight b end cell equals 0 row cell 4 straight a plus straight b end cell equals cell negative 4 space space space... space left parenthesis 1 right parenthesis end cell end table 

Dengan menerapkan dalil L’Hospital yaitu:

limit as x rightwards arrow k of space fraction numerator f left parenthesis x right parenthesis over denominator g left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow k of space fraction numerator f apostrophe left parenthesis x right parenthesis over denominator g apostrophe left parenthesis x right parenthesis end fraction equals limit as x rightwards arrow k of space fraction numerator f double apostrophe left parenthesis x right parenthesis over denominator g double apostrophe left parenthesis x right parenthesis end fraction

Didapat:

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 2 of fraction numerator straight x squared plus 2 ax space plus straight b over denominator straight x minus 2 end fraction end cell equals cell negative 3 end cell row cell limit as straight x rightwards arrow 2 of fraction numerator 2 straight x plus 2 straight a over denominator 1 end fraction end cell equals cell negative 3 end cell row cell limit as straight x rightwards arrow 2 of 2 straight x plus 2 straight a end cell equals cell negative 3 end cell row cell 2 times 2 plus 2 straight a end cell equals cell negative 3 end cell row cell 4 plus 2 straight a end cell equals cell negative 3 end cell row cell 2 straight a end cell equals cell negative 7 end cell row straight a equals cell negative 7 over 2 space space... space left parenthesis 2 right parenthesis end cell end table 

Dari persamaan 1 dan persamaan 2 diperoleh:

Error converting from MathML to accessible text. 

Nilai begin mathsize 14px style straight a times straight b end style:

table attributes columnalign right center left columnspacing 0px end attributes row cell straight a times straight b end cell equals cell negative 7 over 2 times 10 end cell row blank equals cell negative 7 times 5 end cell row blank equals cell negative 35 end cell end table 

Jadi, nilai begin mathsize 14px style straight a times straight b end style adalah negative 35.

Roboguru

Pembahasan Soal:

Diketahui space limit as x rightwards arrow 3 of fraction numerator x squared plus x over denominator 3 end fraction, maka gunakan metode substitusi langsung.

table attributes columnalign right center left columnspacing 0px end attributes row cell space limit as x rightwards arrow 3 of fraction numerator x squared plus x over denominator 3 end fraction end cell equals cell fraction numerator 3 squared plus 3 over denominator 3 end fraction end cell row blank equals cell 12 over 3 end cell row blank equals 4 end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 3 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator x squared plus x over denominator 3 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 4 end table.

Jadi, jawaban yang tepat adalah C.

Roboguru

Pembahasan Soal:

Perhatikan perhitungan berikut.

Ingat:

begin mathsize 14px style 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 equals fraction numerator limit as x rightwards arrow c of f open parentheses x close parentheses over denominator limit as x rightwards arrow c of g open parentheses x close parentheses end fraction end style

Maka:

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 left parenthesis x to the power of 10 minus 1 right parenthesis over denominator 1 minus x squared end fraction end cell equals cell fraction numerator open parentheses 2 to the power of 10 minus 1 close parentheses over denominator 1 minus 2 squared end fraction end cell row blank equals cell fraction numerator 1.024 minus 1 over denominator 1 minus 4 end fraction end cell row blank equals cell fraction numerator 1.023 over denominator negative 3 end fraction end cell row blank equals cell negative 341 end cell end table end style

Jadi, nilai limit tersebut adalah mendekati begin mathsize 14px style negative 341 end style.

Dengan demikian, jawaban yang tepat adalah A.

Roboguru

Tentukan nilai limit fungsi berikut.

Pembahasan Soal:

Diketahui:

Pembagian limit fungsi.

Substitusi variabel begin mathsize 14px style x end style dengan begin mathsize 14px style 1 end style pada fungsi seperti berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 1 of space fraction numerator 3 x cubed minus 4 x squared plus 10 x over denominator 5 x end fraction end cell equals cell fraction numerator 3 open parentheses 1 close parentheses cubed minus 4 open parentheses 1 close parentheses squared plus 10 open parentheses 1 close parentheses over denominator 5 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator 3 minus 4 plus 10 over denominator 5 end fraction end cell row blank equals cell 9 over 5 end cell end table end style

Maka, nilai dari begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell limit as x rightwards arrow 1 of end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 x cubed minus 4 x squared plus 10 x over denominator 5 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 cell 9 over 5 end cell end table end style.

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