x→9limx−3x−9

Pertanyaan

$begin mathsize 14px style limit as x rightwards arrow 9 of fraction numerator x minus 9 over denominator square root of x minus 3 end fraction end style$

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

jawabannya adalah 6.

Pembahasan

Untuk menyelesaikannya rasionalkan penyebutnya terlebih dahulu dengan mengalikannya dengan akar sekawan. Sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 9 of fraction numerator x minus 9 over denominator square root of x minus 3 end fraction end cell equals cell limit as x rightwards arrow 9 of fraction numerator x minus 9 over denominator square root of x minus 3 end fraction cross times fraction numerator square root of x plus 3 over denominator square root of x plus 3 end fraction end cell row blank equals cell limit as x rightwards arrow 9 of fraction numerator up diagonal strike open parentheses x minus 9 close parentheses end strike. open parentheses square root of x plus 3 close parentheses over denominator up diagonal strike x minus 9 end strike end fraction end cell row blank equals cell limit as x rightwards arrow 9 of square root of x plus 3 end cell row blank equals cell square root of 9 plus 3 end cell row blank equals cell 3 plus 3 end cell row blank equals 6 end table$

Jadi, jawabannya adalah 6.

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