Pertanyaan

Diketahui , maka nilai dari 2 p − q = ....

Diketahui $begin mathsize 14px style lim subscript straight x rightwards arrow 2 end subscript fraction numerator square root of straight p open parentheses straight x minus 2 close parentheses plus straight q end root minus 5 over denominator straight x minus 2 end fraction equals negative 2 over 5 end style$ , maka nilai dari $2 p - q =$ ....

R. RGFLSATU

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Perhatikan bahwa jika dilakukan substitusi x = 2 , maka didapatkan hasil Didapat bentuk limit yang jika dilakukan substitusi x = 2 , maka penyebutnya bernilai 0. Tetapi, nilai limitnya ada. Maka haruslah pembilangnya juga bernilai 0 jika disubstitusi x = 2 . Sehingga Maka didapat bahwa Sehingga Jadi, jawaban yang tepat adalah E.

Perhatikan bahwa jika dilakukan substitusi x = 2, maka didapatkan hasil

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator square root of straight p open parentheses straight x minus 2 close parentheses plus straight q end root minus 5 over denominator straight x minus 2 end fraction end cell equals cell fraction numerator square root of straight p open parentheses 2 minus 2 close parentheses plus straight q end root minus 5 over denominator 2 minus 2 end fraction end cell row blank equals cell fraction numerator square root of straight p open parentheses 0 close parentheses plus straight q end root minus 5 over denominator 0 end fraction end cell row blank equals cell fraction numerator square root of straight q minus 5 over denominator 0 end fraction end cell end table end style$

Didapat bentuk limit yang jika dilakukan substitusi x = 2, maka penyebutnya bernilai 0. Tetapi, nilai limitnya ada.
Maka haruslah pembilangnya juga bernilai 0 jika disubstitusi x = 2. Sehingga
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell square root of straight q minus 5 end cell equals 0 row cell square root of straight q end cell equals 5 row straight q equals 25 end table end style

Maka didapat bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell limit as straight x rightwards arrow 2 of fraction numerator square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root minus 5 over denominator straight x minus 2 end fraction end cell equals cell negative 2 over 5 end cell row cell limit as straight x rightwards arrow 2 of open parentheses fraction numerator square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root minus 5 over denominator straight x minus 2 end fraction times fraction numerator square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root plus 5 over denominator square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root plus 5 end fraction close parentheses end cell equals cell negative 2 over 5 end cell row cell limit as straight x rightwards arrow 2 of fraction numerator open parentheses straight p open parentheses straight x minus 2 close parentheses plus 25 close parentheses minus 25 over denominator open parentheses straight x minus 2 close parentheses open parentheses square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root plus 5 close parentheses end fraction end cell equals cell negative 2 over 5 end cell row cell limit as straight x rightwards arrow 2 of fraction numerator straight p open parentheses straight x minus 2 close parentheses over denominator open parentheses straight x minus 2 close parentheses open parentheses square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root plus 5 close parentheses end fraction end cell equals cell negative 2 over 5 end cell row cell limit as straight x rightwards arrow 2 of fraction numerator straight p over denominator square root of straight p open parentheses straight x minus 2 close parentheses plus 25 end root plus 5 end fraction end cell equals cell negative 2 over 5 end cell row cell fraction numerator straight p over denominator square root of straight p open parentheses 2 minus 2 close parentheses plus 25 end root plus 5 end fraction end cell equals cell negative 2 over 5 end cell row cell fraction numerator straight p over denominator square root of straight p open parentheses 0 close parentheses plus 25 end root plus 5 end fraction end cell equals cell negative 2 over 5 end cell row cell fraction numerator straight p over denominator square root of 25 plus 5 end fraction end cell equals cell negative 2 over 5 end cell row cell fraction numerator straight p over denominator 5 plus 5 end fraction end cell equals cell negative 2 over 5 end cell row cell straight p over 10 end cell equals cell negative 2 over 5 end cell row straight p equals cell negative 2 over 5 open parentheses 10 close parentheses end cell row straight p equals cell negative 4 end cell end table end style$

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 straight p minus straight q end cell equals cell 2 open parentheses negative 4 close parentheses minus 25 end cell row blank equals cell negative 8 minus 25 end cell row blank equals cell negative 33 end cell end table end style

Jadi, jawaban yang tepat adalah E.

Latihan Bab

Konsep Kilat

Konsep Limit

Sifat Limit

Limit Fungsi Aljabar