Question

find the horizontal asymptote of ( f(x)=\frac{-3 x^{5}-4 x^{3}+5 x^{2}+2 x}{2 x^{5}+x^{4}-3 x} ). if the horizontal asymptote does not exist, enter dne. the horizontal asymptote is ( y= )

Explanation:

Step1: Divide numerator and denominator by \(x^5\)

$$\begin{align*} \lim_{x ightarrow\pm\infty}\frac{-3x^{5}-4x^{3}+5x^{2}+2x}{2x^{5}+x^{4}-3x}&=\lim_{x ightarrow\pm\infty}\frac{-3-\frac{4}{x^{2}}+\frac{5}{x^{3}}+\frac{2}{x^{4}}}{2+\frac{1}{x}-\frac{3}{x^{4}}}\\ \end{align*}$$

Step2: Evaluate the limit

As \(x
ightarrow\pm\infty\), \(\frac{1}{x}
ightarrow0\), \(\frac{1}{x^{2}}
ightarrow0\), \(\frac{1}{x^{3}}
ightarrow0\), \(\frac{1}{x^{4}}
ightarrow0\)

So \(\lim_{x
ightarrow\pm\infty}\frac{-3-\frac{4}{x^{2}}+\frac{5}{x^{3}}+\frac{2}{x^{4}}}{2+\frac{1}{x}-\frac{3}{x^{4}}}=\frac{-3 - 0+0 + 0}{2+0 - 0}\)

Answer:

\(-\frac{3}{2}\)