## University Calculus: Early Transcendentals (3rd Edition)

$\lim_{x\to-\infty}f(x)/g(x)=2$
Since $f(x)$ and $g(x)$ are polynominals, I would suppose that they have this form: $$f(x)=ax^u+bx^v+...+cx^z$$ $$g(x)=dx^u+ex^v+...+fx^z$$ with $u\gt v\gt...\gt z$ So we would have $$\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=\lim_{x\to\pm\infty}\frac{ax^u+bx^v+...+cx^z}{dx^u+ex^v+...+fx^z}$$ We apply our usual method, which is to divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x^u$ in this case. Notice that the operations until now are applicable to both $x\to\infty$ and $x\to-infty$ without any result difference. $$\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=\lim_{x\to\pm\infty}\frac{a+\frac{b}{x^{u-v}}+...+\frac{c}{x^{u-z}}}{d+\frac{e}{x^{u-v}}+...+\frac{f}{x^{u-z}}}$$ And since $\lim_{x\to\infty}\frac{A}{x^a}=\lim_{x\to-\infty}\frac{A}{x^a}=0$, we have $$\lim_{x\to\pm\infty}\frac{f(x)}{g(x)}=\frac{a+0+...+0}{d+0+...+0}=\frac{a}{d}$$ So, as you can see, the results of limits as $x\to\infty$ and $x\to-\infty$ of $f(x)/g(x)$ are the same. Therefore, if $\lim_{x\to\infty}f(x)/g(x)=2$, then $\lim_{x\to-\infty}f(x)/g(x)=2$