find the limit, if it exists.
51) $lim_{x
ightarrowinfty}\frac{x^{2}-4x + 14}{x^{3}+9x^{2}+5}$
52) $lim_{x
ightarrow-infty}\frac{-8x^{2}+9x + 5}{-15x^{2}+7x + 14}$
53) $lim_{x
ightarrowinfty}\frac{3x + 1}{11x - 7}$
54) $lim_{x
ightarrowinfty}\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}$

Question

Accepted Answer

### 51)
# Explanation:
## 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 $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{3}}=0$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{\frac{1}{x}-\frac{4}{x^{2}}+\frac{14}{x^{3}}}{1+\frac{9}{x}+\frac{5}{x^{3}}}&=\frac{0 - 0+0}{1 + 0+0}\
&=0
\end{align*}
$$
# Answer:
$0$

### 52)
# Explanation:
## 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 $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}} = 0$.
$$
\begin{align*}
\lim_{x
ightarrow-\infty}\frac{-8+\frac{9}{x}+\frac{5}{x^{2}}}{-15+\frac{7}{x}+\frac{14}{x^{2}}}&=\frac{-8 + 0+0}{-15+0+0}\
&=\frac{8}{15}
\end{align*}
$$
# Answer:
$\frac{8}{15}$

### 53)
# Explanation:
## 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:
$\frac{3}{11}$

### 54)
# Explanation:
## Step1: Divide by highest - power of x in denominator
Divide numerator and denominator by $x^{3}$:
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{7x^{3}-3x^{2}+3x}{-x^{3}-2x + 7}&=\lim_{x
ightarrow\infty}\frac{\frac{7x^{3}}{x^{3}}-\frac{3x^{2}}{x^{3}}+\frac{3x}{x^{3}}}{\frac{-x^{3}}{x^{3}}-\frac{2x}{x^{3}}+\frac{7}{x^{3}}}\
&=\lim_{x
ightarrow\infty}\frac{7-\frac{3}{x}+\frac{3}{x^{2}}}{-1-\frac{2}{x^{2}}+\frac{7}{x^{3}}}
\end{align*}
$$
## Step2: Use limit rules for infinity
As $x
ightarrow\infty$, $\lim_{x
ightarrow\infty}\frac{1}{x}=0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$, $\lim_{x
ightarrow\infty}\frac{1}{x^{3}}=0$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{7-\frac{3}{x}+\frac{3}{x^{2}}}{-1-\frac{2}{x^{2}}+\frac{7}{x^{3}}}&=\frac{7-0 + 0}{-1-0+0}\
&=-7
\end{align*}
$$
# Answer:
$-7$

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QUESTION IMAGE

Question

Explanation:

51)

Step1: Divide by highest - power of x in denominator

Step2: Use limit rules for infinity

Step1: Divide by highest - power of x in denominator

Step2: Use limit rules for infinity

Step1: Divide by highest - power of x in denominator

Step2: Use limit rules for infinity

Answer:

52)