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Jika begin mathsize 14px style limit as x rightwards arrow 4 of fraction numerator a x plus b minus square root of x over denominator x minus 4 end fraction equals 3 over 4 comma space maka space straight a plus straight b equals... end style

Pembahasan Soal:

Langkah awal menyelesaikan limit adalah dengan mensubstitusi nilai x:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator a x plus b minus square root of x over denominator x minus 4 end fraction end cell equals cell fraction numerator 4 a plus b minus square root of 4 over denominator 4 minus 4 end fraction end cell row blank equals cell fraction numerator 4 a plus b minus 2 over denominator 0 end fraction end cell end table end style

Agar limit tersebut mempunyai nilai, maka hasil substitusi nya harus lah begin mathsize 14px style 0 over 0 end style, sehingga begin mathsize 14px style 4 a plus b minus 2 equals 0 end style. Untuk menyelesaikannya, gunakan dalil l'hospital yaitu

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

Sehingga diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 4 of fraction numerator a x plus b minus square root of x over denominator x minus 4 end fraction end cell equals cell limit as x rightwards arrow 4 of fraction numerator a minus begin display style fraction numerator 1 over denominator 2 square root of x end fraction end style over denominator 1 end fraction end cell row blank equals cell a minus fraction numerator 1 over denominator 2 square root of 4 end fraction end cell row blank equals cell a minus 1 fourth end cell end table end style

Pada soal, diketahui nilai limit nya adalah begin mathsize 14px style 3 over 4 end style, maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a minus 1 fourth end cell equals cell 3 over 4 end cell row a equals 1 end table end style

Kemudian substitusikan nilai a ke persamaan begin mathsize 14px style 4 a plus b minus 2 equals 0 end style diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 4 left parenthesis 1 right parenthesis plus b minus 2 end cell equals 0 row b equals cell negative 2 end cell end table end style

Jadi, nilai begin mathsize 14px style a plus b equals 1 plus open parentheses negative 2 close parentheses equals negative 1 end style.

 

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 30 Juli 2021

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