Answer
$\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
Work Step by Step
$\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