## Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to-\infty}\frac{1+x^6}{x^4+1}=\infty$$
$$A=\lim\limits_{x\to-\infty}\frac{1+x^6}{x^4+1}$$$$=\lim\limits_{x\to-\infty}\frac{X}{Y}$$ Divide both numerator and denominator by $x^4$ (the highest-power variable in the denominator), we have - In the numerator: $X=\frac{1+x^6}{x^4}=\frac{1}{x^4}+x^2$ - In the denominator: $Y=\frac{x^4+1}{x^4}=1+\frac{1}{x^4}$ Therefore, $$A=\lim\limits_{x\to-\infty}\frac{\frac{1}{x^4}+x^2}{1+\frac{1}{x^4}}$$ As $x\to-\infty$, $\frac{1}{x^4}+x^2$ approaches $0+\infty=\infty$ while $1+\frac{1}{x^4}$ approaches $1+0=1$ Therefore, $$A=\infty$$