Answer
$\lim\limits_{x\to-\infty}\frac{x-2}{x^2+1}=0$
Work Step by Step
$$A=\lim\limits_{x\to-\infty}\frac{x-2}{x^2+1}$$
Divide both numerator and denominator by $x^2$
$A=\lim\limits_{x\to-\infty}\frac{\frac{x-2}{x^2}}{\frac{x^2+1}{x^2}}$
$A=\frac{\lim\limits_{x\to-\infty}\frac{x-2}{x^2}}{\lim\limits_{x\to-\infty}\frac{x^2+1}{x^2}}$
$A=\frac{\lim\limits_{x\to-\infty}(\frac{1}{x}-\frac{2}{x^2})}{\lim\limits_{x\to-\infty}(1+\frac{1}{x^2})}$
$A=\frac{\lim\limits_{x\to-\infty}\frac{1}{x}-\lim\limits_{x\to-\infty}\frac{2}{x^2}}{\lim\limits_{x\to-\infty}1+\lim\limits_{x\to-\infty}\frac{1}{x^2}}$
$A=\frac{0-2\times0}{1+0}$
$A=0$
