## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to-\infty}\frac{1-2x^2-x^4}{5+x-3x^4}=\frac{1}{3}$$
$$A=\lim\limits_{x\to-\infty}\frac{1-2x^2-x^4}{5+x-3x^4}$$ Divide both numerator and denominator by $x^4$ (the highest power of $x$ in the denominator), we have: - Numerator: $$\frac{1-2x^2-x^4}{x^4}=\frac{1}{x^4}-\frac{2}{x^2}-1$$ - Denominator: $$\frac{5+x-3x^4}{x^4}=\frac{5}{x^4}+\frac{1}{x^3}-3$$ Therefore, $$A=\lim\limits_{x\to-\infty}\frac{\frac{1}{x^4}-\frac{2}{x^2}-1}{\frac{5}{x^4}+\frac{1}{x^3}-3}$$$$A=\frac{0-2\times0-1}{5\times0+0-3}$$$$A=\frac{1}{3}$$