Pertanyaan

Tentukan nilai a ⋅ b . x → 2 lim x − 2 x 2 + 2 ax + b = − 3

Tentukan nilai $a \cdot b$ .

$x \to 2 lim \frac{x ^{2} + 2 ax + b}{x - 2} = - 3$

E. Dwi

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

nilai adalah .

nilai

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

adalah

Pembahasan

Berdasarkan sifat limit yaitu: Dengan menggunakan cara substitusi, substitusikannilai : Karena dengan mensubstitusikan limit tersebut tidak terdefinisi.maka menggunakan cara lain untuk menyelesaikan permasalahan di atas. mempunyai nilai jika untuk , Maka diperoleh: Dengan menerapkandalil L’Hospital yaitu: Didapat: Dari persamaan 1 dan persamaan 2 diperoleh: Nilai : Jadi, nilai adalah .

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 untuk , 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:

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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 .