Question

find the limit, if it exists.

1. $lim_{x

ightarrowinfty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}$

1. $lim_{x

ightarrow-infty}\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}$

1. $lim_{x

ightarrowinfty}\frac{3x + 1}{11x - 7}$

1. $lim_{x

ightarrowinfty}\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}$

Explanation:

51)

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x^{3}$:
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}&=\lim_{x ightarrow\infty}\frac{\frac{x^{2}}{x^{3}}-\frac{4x}{x^{3}}+\frac{14}{x^{3}}}{\frac{x^{3}}{x^{3}}+\frac{9x^{2}}{x^{3}}+\frac{5}{x^{3}}}\\ &=\lim_{x ightarrow\infty}\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1 + \frac{9}{x}+\frac{5}{x^{3}}} \end{align*}$$

\]

Step2: Use limit rules for infinity

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{3}}=0$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1+\frac{9}{x}+\frac{5}{x^{3}}}&=\frac{0 - 0+0}{1 + 0+0}\\ &=0 \end{align*}$$

\]

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x^{2}$:
\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}&=\lim_{x ightarrow-\infty}\frac{\frac{-8x^{2}}{x^{2}}+\frac{9x}{x^{2}}+\frac{5}{x^{2}}}{\frac{-15x^{2}}{x^{2}}+\frac{7x}{x^{2}}+\frac{14}{x^{2}}}\\ &=\lim_{x ightarrow-\infty}\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}} \end{align*}$$

\]

Step2: Use limit rules for infinity

As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}} = 0$.
\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}}&=\frac{-8 + 0+0}{-15+0+0}\\ &=\frac{8}{15} \end{align*}$$

\]

Step1: Divide by highest - power of x in denominator

Divide numerator and denominator by $x$:
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{3x + 1}{11x-7}&=\lim_{x ightarrow\infty}\frac{\frac{3x}{x}+\frac{1}{x}}{\frac{11x}{x}-\frac{7}{x}}\\ &=\lim_{x ightarrow\infty}\frac{3+\frac{1}{x}}{11-\frac{7}{x}} \end{align*}$$

\]

Step2: Use limit rules for infinity

As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{3+\frac{1}{x}}{11-\frac{7}{x}}&=\frac{3 + 0}{11-0}\\ &=\frac{3}{11} \end{align*}$$

\]

Answer:

$0$

52)