Question

mptote, a slant asymptote, or neither. then

1. $h(x)=\frac{x^{3}-2x + 5}{3x - 4}$ $\frac{x^{3}}{3x}$ neither.

$lim_{x
ightarrow-infty}h(x)=$
$lim_{x
ightarrowinfty}h(x)=$

1. $y=\frac{(x - 2)^{2}(3x^{2}+2)}{x(x + 1)(x - 5)}$ $\frac{x^{5}}{x^{3}}$ neither.

$lim_{x
ightarrow-infty}y=$
$lim_{x
ightarrowinfty}y=$

Explanation:

Step1: Analyze degree of numerator and denominator for $h(x)$

For $h(x)=\frac{x^{3}-2x + 5}{3x-4}$, the degree of the numerator $n = 3$ and the degree of the denominator $m=1$. Since $n>m$, we use long - division.
$$\frac{x^{3}-2x + 5}{3x-4}=\frac{1}{3}x^{2}+\frac{4}{9}x+\frac{10}{27}+\frac{173}{27(3x - 4)}$$
As $x\to\pm\infty$, $\lim_{x\to\pm\infty}h(x)=\lim_{x\to\pm\infty}(\frac{1}{3}x^{2}+\frac{4}{9}x+\frac{10}{27}+\frac{173}{27(3x - 4)})=\infty$.

Step2: Analyze degree of numerator and denominator for $y$

For $y=\frac{(x - 2)^{2}(3x^{2}+2)}{x(x + 1)(x - 5)}$, expand the numerator: $(x - 2)^{2}(3x^{2}+2)=(x^{2}-4x + 4)(3x^{2}+2)=3x^{4}-12x^{3}+14x^{2}-8x + 8$.
Expand the denominator: $x(x + 1)(x - 5)=x(x^{2}-4x-5)=x^{3}-4x^{2}-5x$.
The degree of the numerator $n = 4$ and the degree of the denominator $m = 3$. Using long - division, we find that as $x\to\pm\infty$, $\lim_{x\to\pm\infty}y=\infty$.

Step3: Find limits for $h(x)$

For $h(x)=\frac{x^{3}-2x + 5}{3x-4}$,
$\lim_{x\to-\infty}h(x)=-\infty$ and $\lim_{x\to\infty}h(x)=\infty$

Step4: Find limits for $y$

For $y=\frac{(x - 2)^{2}(3x^{2}+2)}{x(x + 1)(x - 5)}$,
$\lim_{x\to-\infty}y=-\infty$ and $\lim_{x\to\infty}y=\infty$

Answer:

For $h(x)=\frac{x^{3}-2x + 5}{3x-4}$, $\lim_{x\to-\infty}h(x)=-\infty$, $\lim_{x\to\infty}h(x)=\infty$
For $y=\frac{(x - 2)^{2}(3x^{2}+2)}{x(x + 1)(x - 5)}$, $\lim_{x\to-\infty}y=-\infty$, $\lim_{x\to\infty}y=\infty$