question 4\nfor each function, determine the long - run behavior\n$\frac{x^{3}+1}{x^{2}+2}$ has a horizontal…

question 4\nfor each function, determine the long - run behavior\n$\frac{x^{3}+1}{x^{2}+2}$ has a horizontal asymptote at y = 1\n$\frac{x^{2}+1}{x^{3}+2}$ has a horizontal asymptote at y = 0\n$\frac{x^{2}+1}{x^{2}+2}$ has no horizontal asymptote\nhide work entry\nshow work here by typing it or attaching a file or picture\nedit insert paragraph\nboth numerator and denominator are degree 2. leading coefficients are both 1, so the long - run behavior is their ratio.\n$lim_{x\rightarrowpminfty}\frac{x^{2}+1}{x^{2}+2}=1$ $lim_{x\rightarrowpminfty}\frac{x^{2}+2}{x^{2}+1}=1$\nanswer: 1\nquestion 2\n$x^{2}+1x^{3}+2x^{3}+2x^{2}+1$\nnumerator degree 2, denominator degree 3. denominator grows faster, so the fraction goes to 0.\n$lim_{x\rightarrowpminfty}\frac{x^{2}+1}{x^{3}+2}=0$ $lim_{x\rightarrowpminfty}\frac{x^{3}+2}{x^{2}+1}=infty$\nnext question

question 4\nfor each function, determine the long - run behavior\n$\frac{x^{3}+1}{x^{2}+2}$ has a horizontal asymptote at y = 1\n$\frac{x^{2}+1}{x^{3}+2}$ has a horizontal asymptote at y = 0\n$\frac{x^{2}+1}{x^{2}+2}$ has no horizontal asymptote\nhide work entry\nshow work here by typing it or attaching a file or picture\nedit insert paragraph\nboth numerator and denominator are degree 2. leading coefficients are both 1, so the long - run behavior is their ratio.\n$lim_{x\rightarrowpminfty}\frac{x^{2}+1}{x^{2}+2}=1$ $lim_{x\rightarrowpminfty}\frac{x^{2}+2}{x^{2}+1}=1$\nanswer: 1\nquestion 2\n$x^{2}+1x^{3}+2x^{3}+2x^{2}+1$\nnumerator degree 2, denominator degree 3. denominator grows faster, so the fraction goes to 0.\n$lim_{x\rightarrowpminfty}\frac{x^{2}+1}{x^{3}+2}=0$ $lim_{x\rightarrowpminfty}\frac{x^{3}+2}{x^{2}+1}=infty$\nnext question

Answer

Explanation:

Step1: Recall horizontal - asymptote rules

For a rational function $y = \frac{f(x)}{g(x)}=\frac{a_nx^n+\cdots+a_0}{b_mx^m+\cdots + b_0}$, if $n = m$, the horizontal asymptote is $y=\frac{a_n}{b_m}$; if $n<m$, the horizontal asymptote is $y = 0$; if $n>m$, there is no horizontal asymptote.

Step2: Analyze $\frac{x^3 + 1}{x^2+2}$

The degree of the numerator $n = 3$ and the degree of the denominator $m = 2$. Since $n>m$, there is no horizontal asymptote.

Step3: Analyze $\frac{x^2 + 1}{x^3+2}$

The degree of the numerator $n = 2$ and the degree of the denominator $m = 3$. Since $n<m$, $\lim_{x\rightarrow\pm\infty}\frac{x^2 + 1}{x^3+2}=0$, so the horizontal asymptote is $y = 0$.

Step4: Analyze $\frac{x^2 + 1}{x^2+2}$

The degree of the numerator $n = 2$ and the degree of the denominator $m = 2$. The leading - coefficient of the numerator $a_n=1$ and the leading - coefficient of the denominator $b_m = 1$. Then $\lim_{x\rightarrow\pm\infty}\frac{x^2 + 1}{x^2+2}=\lim_{x\rightarrow\pm\infty}\frac{1+\frac{1}{x^2}}{1+\frac{2}{x^2}} = 1$, so the horizontal asymptote is $y = 1$.

Answer:

  • $\frac{x^3 + 1}{x^2+2}$: No horizontal asymptote
  • $\frac{x^2 + 1}{x^3+2}$: a Horizontal asymptote at $y = 0$
  • $\frac{x^2 + 1}{x^2+2}$: a Horizontal asymptote at $y = 1$