Question

for each function, determine the long run behavior
\\( \frac { x ^ { 3 } + 1 } { x ^ { 2 } + 2 } \\) has
\\( \frac { x ^ { 2 } + 1 } { x ^ { 3 } + 2 } \\) has
\\( \frac { x ^ { 2 } + 1 } { x ^ { 2 } + 2 } \\) has

Explanation:

Step1: Analyze the degrees of numerator and denominator for \(\frac{x^{3}+1}{x^{2}+2}\)

For a rational function \(\frac{f(x)}{g(x)}\), if \(\text{deg}(f(x))>\text{deg}(g(x))\), there is no horizontal asymptote. Here, \(\text{deg}(x^{3}+1) = 3\) and \(\text{deg}(x^{2}+2)=2\).

Step2: Analyze the degrees of numerator and denominator for \(\frac{x^{2}+1}{x^{3}+2}\)

If \(\text{deg}(f(x))<\text{deg}(g(x))\), the horizontal asymptote is \(y = 0\). Here, \(\text{deg}(x^{2}+1)=2\) and \(\text{deg}(x^{3}+2) = 3\).

Step3: Analyze the degrees of numerator and denominator for \(\frac{x^{2}+1}{x^{2}+2}\)

If \(\text{deg}(f(x))=\text{deg}(g(x))\), the horizontal asymptote is \(y=\frac{a_{n}}{b_{n}}\), where \(a_{n}\) and \(b_{n}\) are the leading coefficients of \(f(x)\) and \(g(x)\). For \(f(x)=x^{2}+1\) (leading coefficient \(a_{n}=1\)) and \(g(x)=x^{2}+2\) (leading coefficient \(b_{n}=1\)), \(y = 1\).

Answer:

\(\frac{x^{3}+1}{x^{2}+2}\) has No horizontal asymptote.
\(\frac{x^{2}+1}{x^{3}+2}\) has a Horizontal asymptote at \(y = 0\).
\(\frac{x^{2}+1}{x^{2}+2}\) has a Horizontal asymptote at \(y = 1\).