Pertanyaan

...

$begin mathsize 14px style lim subscript straight x rightwards arrow negative 1 end subscript fraction numerator cube root of straight x squared end root plus 2 cube root of straight x plus 1 over denominator open parentheses straight x plus 1 close parentheses squared end fraction equals end style$ ...

D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Perhatikan bahwa jika dilakukan substitusi x = -1 , maka didapatkan hasil . Sehingga perlu dicari cara lain untuk mencari nilai limitnya. Misalkan Maka untuk , maka didapat bahwa . Sehingga

Perhatikan bahwa jika dilakukan substitusi x = -1, maka didapatkan hasil .
Sehingga perlu dicari cara lain untuk mencari nilai limitnya.

Misalkan
Maka untuk , maka didapat bahwa .
Sehingga

$begin mathsize 14px style limit as x rightwards arrow negative 1 of fraction numerator cube root of x squared end root plus 2 cube root of x plus 1 over denominator open parentheses x plus 1 close parentheses squared end fraction equals limit as y rightwards arrow negative 1 of fraction numerator cube root of open parentheses y cubed close parentheses squared end root plus 2 cube root of y cubed end root plus 1 over denominator open parentheses y cubed plus 1 close parentheses squared end fraction equals limit as y rightwards arrow negative 1 of fraction numerator y squared plus 2 y plus 1 over denominator open parentheses y cubed plus 1 close parentheses squared end fraction equals limit as y rightwards arrow negative 1 of open parentheses y plus 1 close parentheses squared over open parentheses open parentheses y plus 1 close parentheses open parentheses y squared minus y plus 1 close parentheses close parentheses squared equals limit as y rightwards arrow negative 1 of fraction numerator open parentheses y plus 1 close parentheses squared over denominator open parentheses y plus 1 close parentheses squared open parentheses y squared minus y plus 1 close parentheses squared end fraction equals limit as y rightwards arrow negative 1 of 1 over open parentheses y squared minus y plus 1 close parentheses squared equals 1 over open parentheses open parentheses negative 1 close parentheses squared minus open parentheses negative 1 close parentheses plus 1 close parentheses squared equals 1 over open parentheses 1 plus 1 plus 1 close parentheses squared equals 1 over open parentheses 3 close parentheses squared equals 1 over 9 end style$