Question

for each function, determine the long run behavior.
\frac{x^{2}+1}{x^{3}+2} has select an answer
\frac{x^{2}+1}{x^{2}+2} has select an answer
\frac{x^{3}+1}{x^{2}+2} has select an answer
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Explanation:

Step1: Divide numerator and denominator by highest - power of \(x\) in denominator

For \(\frac{x^{2}+1}{x^{3}+2}\), divide numerator and denominator by \(x^{3}\):

$$ \frac{\frac{x^{2}}{x^{3}}+\frac{1}{x^{3}}}{\frac{x^{3}}{x^{3}}+\frac{2}{x^{3}}}=\frac{\frac{1}{x}+\frac{1}{x^{3}}}{1 + \frac{2}{x^{3}}} $$

As \(x\to\pm\infty\), \(\frac{1}{x}\to0\) and \(\frac{1}{x^{3}}\to0\). So \(\lim_{x\to\pm\infty}\frac{x^{2}+1}{x^{3}+2}=0\)

Step2: Divide numerator and denominator by highest - power of \(x\) in denominator

For \(\frac{x^{2}+1}{x^{2}+2}\), divide numerator and denominator by \(x^{2}\):

$$ \frac{\frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{2}{x^{2}}}=\frac{1+\frac{1}{x^{2}}}{1+\frac{2}{x^{2}}} $$

As \(x\to\pm\infty\), \(\frac{1}{x^{2}}\to0\). So \(\lim_{x\to\pm\infty}\frac{x^{2}+1}{x^{2}+2}=1\)

Step3: Divide numerator and denominator by highest - power of \(x\) in denominator

For \(\frac{x^{3}+1}{x^{2}+2}\), divide numerator and denominator by \(x^{2}\):

$$ \frac{\frac{x^{3}}{x^{2}}+\frac{1}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{2}{x^{2}}}=\frac{x+\frac{1}{x^{2}}}{1+\frac{2}{x^{2}}} $$

As \(x\to\infty\), \(\frac{x+\frac{1}{x^{2}}}{1+\frac{2}{x^{2}}}\to\infty\); as \(x\to-\infty\), \(\frac{x+\frac{1}{x^{2}}}{1+\frac{2}{x^{2}}}\to-\infty\)

Answer:

\(\frac{x^{2}+1}{x^{3}+2}\) has a long - run behavior of approaching \(0\); \(\frac{x^{2}+1}{x^{2}+2}\) has a long - run behavior of approaching \(1\); \(\frac{x^{3}+1}{x^{2}+2}\) has a long - run behavior of approaching \(\pm\infty\) (depending on the direction of \(x\) approaching infinity)