Tentukan nilai dari limit berikut
[tex]\displaystyle \lim_{x\to 1}\left [ \frac{1}{2(1-x^3)}-\frac{1}{3(1-x^2)} \right ][/tex]
[tex]\displaystyle \lim_{x\to 1}\left [ \frac{1}{2(1-x^3)}-\frac{1}{3(1-x^2)} \right ][/tex]
Verified answer
Jawab:
[tex]\displaystyle=\frac{1}{12}[/tex]
Penjelasan:
Sederhanakan bagian dalam
[tex]\displaystyle\frac{1}{2\left(1-x^3\right)}-\frac{1}{3\left(1-x^2\right)}\\\\\because a^2-b^2=~~~~~(a-b)(a+b)\\~~~~a^3-b^3=(a-b)(a^2+ab+b^2)\therefore\\\\=\frac{1}{2\left(1-x\right)\left(1+x+x^2\right)}-\frac{1}{3\left(1-x\right)\left(1+x\right)}\\\\=\frac{3\left(1+x\right)-2(1+x+x^2)}{6\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}\\\\=\frac{3+3x-2-2x-2x^2}{6\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}[/tex]
[tex]\displaystyle =\frac{1+x-2x^2}{6\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}\\\\=\frac{-(1-x+2x^2)}{6\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}\\\\=\frac{-(2x+1)(x-1)}{6\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}\\\\=\frac{(2x+1)\not(1-x)}{6\not\left(1-x\right)\left(1+x\right)\left(1+x+x^2\right)}\\\\=\frac{2x+1}{6(1+x)(1+x+x^2)}\\\\=\frac{2x+1}{6(1+x)(1+x(1+x))}[/tex]
Masukkan nilai x = 1
[tex]\displaystyle=\frac{2(1)+1}{6(1+1)(1+1(1+1))}\\\\=\frac{2+1}{6(2)(1+1(2))}\\\\=\frac{3}{6(2)(1+2)}\\\\=\frac{3}{6(2)(3)}\\\\=\frac{1}{12}[/tex]
(xcvi)