Answer
$\lim\limits_{x\to\infty}\frac{1-x^2}{x^3-x+1}=0$
Work Step by Step
$$A=\lim\limits_{x\to\infty}\frac{1-x^2}{x^3-x+1}$$
Divide both numerator and denominator by $x^3$
$A=\lim\limits_{x\to\infty}\frac{\frac{1-x^2}{x^3}}{\frac{x^3-x+1}{x^3}}$
$A=\lim\limits_{x\to\infty}\frac{\frac{1}{x^3}-\frac{1}{x}}{1-\frac{1}{x^2}+{\frac{1}{x^3}}}$
$A=\frac{\lim\limits_{x\to\infty}(\frac{1}{x^3}-\frac{1}{x})}{\lim\limits_{x\to\infty}(1-\frac{1}{x^2}+\frac{1}{x^3})}$
$A=\frac{\lim\limits_{x\to\infty}\frac{1}{x^3}-\lim\limits_{x\to\infty}\frac{1}{x}}{\lim\limits_{x\to\infty}1-\lim\limits_{x\to\infty}\frac{1}{x^2}+\lim\limits_{x\to\infty}\frac{1}{x^3}}$
$A=\frac{0-0}{1-0+0}$
$A=0$
