## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to\infty}\frac{x+x^2}{1-2x^2}=-\frac{1}{2}$$
$$A=\lim_{x\to\infty}\frac{x+x^2}{1-2x^2}$$ 1) Method 1: Elementary method Divide both numerator and denominator by $x^2$, which is the highest power in the denominator. $$A=\lim_{x\to\infty}\frac{\frac{x+x^2}{x^2}}{\frac{1-2x^2}{x^2}}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{x}+1}{\frac{1}{x^2}-2}$$ $$A=\frac{0+1}{0-2}$$ (since $\lim_{x\to\infty}\frac{1}{x}=\lim_{x\to\infty}\frac{1}{x^2}=0$) $$A=-\frac{1}{2}$$ 2) Method 2: L'Hospital's Rule As $x\to\infty$, $x+x^2$ approaches $\infty$ and $1-2x^2$ approaches $-\infty$. So this limit is in an indeterminate form of $\infty/-\infty$. We can use L'Hospital's Rule here: $$A=\lim_{x\to\infty}\frac{\frac{d}{dx}(x+x^2)}{\frac{d}{dx}(1-2x^2)}$$ $$A=\lim_{x\to\infty}\frac{1+2x}{-4x}$$ Divide both numerator and denominator by $x$, which is the highest power in the denominator, we have $$A=\lim_{x\to\infty}\frac{\frac{1+2x}{x}}{\frac{-4x}{x}}$$ $$A=\lim_{x\to\infty}\frac{\frac{1}{x}+2}{-4}$$ We remember that $\lim_{x\to\infty}(\frac{1}{x})=0$. So, $$A=\frac{0+2}{-4}$$ $$A=-\frac{1}{2}$$