## Calculus: Early Transcendentals (2nd Edition)

$\lim _{x\rightarrow \infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\infty ;\lim _{x\rightarrow -\infty }\dfrac {4x^{5}+x^{2}}{16x^{4}-2x}=-\infty$ Doesn’t have horizontal asymptote
$\lim _{x\rightarrow \infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\dfrac {40x+\dfrac {1}{x^{2}}}{16-\dfrac {2}{x^{3}}}=\dfrac {40x+0}{16-0}=\dfrac {40x}{16}=\dfrac {5}{2}x=\infty$ $\lim _{x\rightarrow -\infty }\dfrac {40x^{5}+x^{2}}{16x^{4}-2x}=\dfrac {40x+\dfrac {1}{x^{2}}}{16-\dfrac {2}{x^{3}}}=\dfrac {40x+0}{16-0}=\dfrac {40x}{16}=\dfrac {5}{2}x=-\infty$ Since both limits are infinite this equation doesn’t have horizontal asymptote