## Calculus: Early Transcendentals (2nd Edition)

$\lim _{x\rightarrow \infty }\dfrac {-x^{3}+1}{2x+8}=-\infty$ $\lim _{x\rightarrow -\infty }\dfrac {-x^{3}+1}{2x+8}=\infty$ Doesn’t have horizontal asymptote
$\lim _{x\rightarrow \infty }\dfrac {-x^{3}+1}{2x+8}=\dfrac {-x^{2}+\dfrac {1}{x}}{2+\dfrac {8}{x}}=\dfrac {-x^{2}+0}{2+0}=-\dfrac {x^{2}}{2}=-\infty$ $\lim _{x\rightarrow -\infty }\dfrac {-x^{3}+1}{2x+8}=\dfrac {-x^{2}+\dfrac {1}{x}}{2+\dfrac {8}{x}}=\dfrac {-x^{2}+0}{2+0}=-\dfrac {x^{2}}{2}=\infty$ Since both limits are infinite this function doesn’t have horizontal asymptote