Question

question 4
for each function, determine the long - run behavior
$\frac{x^{3}+1}{x^{2}+2}$ has a horizontal asymptote at y = 1
$\frac{x^{2}+1}{x^{3}+2}$ has a horizontal asymptote at y = 0
$\frac{x^{2}+1}{x^{2}+2}$ has no horizontal asymptote
hide work entry
show work here by typing it or attaching a file or picture
edit insert paragraph
both numerator and denominator are degree 2. leading coefficients are both 1, so the long - run behavior is their ratio.
$lim_{x
ightarrowpminfty}\frac{x^{2}+1}{x^{2}+2}=1$ $lim_{x
ightarrowpminfty}\frac{x^{2}+2}{x^{2}+1}=1$
answer: 1
question 2
$x^{2}+1x^{3}+2x^{3}+2x^{2}+1$
numerator degree 2, denominator degree 3. denominator grows faster, so the fraction goes to 0.
$lim_{x
ightarrowpminfty}\frac{x^{2}+1}{x^{3}+2}=0$ $lim_{x
ightarrowpminfty}\frac{x^{3}+2}{x^{2}+1}=infty$
next question

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 $nm$, 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 $nightarrow\pm\infty}\frac{x^2 + 1}{x^3+2}=0$, 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 $\lim_{x
ightarrow\pm\infty}\frac{x^2 + 1}{x^2+2}=\lim_{x
ightarrow\pm\infty}\frac{1+\frac{1}{x^2}}{1+\frac{2}{x^2}} = 1$, 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 $y = 1$