## Calculus: Early Transcendentals 8th Edition

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