I am finding Horizontal and vertical asymptotes of function $$x^{-3}e^{\frac{x^3}{3}}$$
and i have litte problem with horizontal, because $$y =kx +q$$ $$\lim_{x \to \infty}{\frac{f(x)}{x}} = k \in R$$ $$\lim_{x \to \infty}{f(x)-kx} = q \in R $$ But first limit equals infinity, what does it mean? Because when i draw graph in WolframAlpha i see thete horizonal asymptote = 0. Thanks for advice.
The function have not asymptote when $x \to +\infty$. The only asymptote is $y=0$ when $x \to -\infty$.
$\displaystyle \lim_{x \to -\infty} x^{-3}e^{\frac{x^3}{3}} = \lim_{x \to -\infty} x^{-3} \times \lim_{x \to -\infty} e^{\frac{x^3}{3}}=0 \times 0 =0$.
The formula $\lim_{x \to \infty}{\frac{f(x)}{x}} = k \in R$ is good also for $k=0$ but if directly $\lim_{x \to \infty}{f(x)} =0$ this is sufficient to conclude that $y=0$ is asymptote in $+\infty$.