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standard 9 homework
score: 4/27 answered: 4/20
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question 5
find the horizontal asymptote of ( f(x)=\frac{3 x+x^{3}-4}{5 x^{3}+4 x^{2}-5} ). if the horizontal asymptote does not exist, enter dne.
the horizontal asymptote is ( y=)
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Explanation:

Step1: Divide numerator and denominator by \(x^3\)

$$\lim_{x ightarrow\pm\infty}\frac{3x + x^3 - 4}{5x^3 + 4x^2 - 5}=\lim_{x ightarrow\pm\infty}\frac{\frac{3x}{x^3}+\frac{x^3}{x^3}-\frac{4}{x^3}}{\frac{5x^3}{x^3}+\frac{4x^2}{x^3}-\frac{5}{x^3}}$$

Step2: Simplify the expression

$$=\lim_{x ightarrow\pm\infty}\frac{\frac{3}{x^2}+1-\frac{4}{x^3}}{5+\frac{4}{x}-\frac{5}{x^3}}$$

Step3: Evaluate the limit

As \(x
ightarrow\pm\infty\), \(\frac{3}{x^2}
ightarrow0\), \(\frac{4}{x^3}
ightarrow0\), \(\frac{4}{x}
ightarrow0\), \(\frac{5}{x^3}
ightarrow0\)
$$=\frac{0 + 1-0}{5+0 - 0}=\frac{1}{5}$$

Answer:

\(\frac{1}{5}\)