55) $lim_{x
ightarrow -infty}\frac{5x^{3}+4x^{2}}{x - 5x^{2}}$
56) $lim_{x
ightarrow-infty}\frac{-6+(2/x)}{7-(1/x^{2})}$
57) $lim_{x
ightarrow-infty}\frac{cos 3x}{x}$
58) $lim_{x
ightarrowinfty}\frac{-5sqrt{x}+x - 1}{-4x + 2}$
59) $lim_{x
ightarrowinfty}\frac{3x^{-1}+-2x^{-3}}{3x^{-2}+x^{-5}}$
60) $lim_{x
ightarrow-infty}\frac{sqrt3{x}+5x+-5}{3x + x^{2/3}+-4}$

Question

Accepted Answer

### 55)
# Explanation:
## Step1: Divide numerator and denominator by highest - power of x in denominator
Divide both the numerator and denominator of $\frac{5x^{3}+4x^{2}}{x - 5x^{2}}$ by $x^{2}$. We get $\lim_{x
ightarrow-\infty}\frac{5x + 4}{\frac{1}{x}-5}$.
## Step2: Evaluate the limit
As $x
ightarrow-\infty$, $\frac{1}{x}
ightarrow0$. So, $\lim_{x
ightarrow-\infty}\frac{5x + 4}{\frac{1}{x}-5}=-\infty$.

# Answer:
$-\infty$

### 56)
# Explanation:
## Step1: Use limit properties
We know that $\lim_{x
ightarrow-\infty}\frac{-6+\frac{2}{x}}{7-\frac{1}{x^{2}}}$. As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{2}{x}=0$ and $\lim_{x
ightarrow-\infty}\frac{1}{x^{2}} = 0$.
## Step2: Substitute the limit values
Substituting these values into the expression, we get $\frac{-6 + 0}{7-0}=-\frac{6}{7}$.

# Answer:
$-\frac{6}{7}$

### 57)
# Explanation:
## Step1: Apply the Squeeze Theorem
We know that $- 1\leqslant\cos(3x)\leqslant1$. Then $\frac{-1}{x}\leqslant\frac{\cos(3x)}{x}\leqslant\frac{1}{x}$.
## Step2: Evaluate the limits of the bounding functions
As $x
ightarrow-\infty$, $\lim_{x
ightarrow-\infty}\frac{-1}{x}=0$ and $\lim_{x
ightarrow-\infty}\frac{1}{x}=0$. By the Squeeze Theorem, $\lim_{x
ightarrow-\infty}\frac{\cos(3x)}{x}=0$.

# Answer:
$0$

### 58)
# Explanation:
## Step1: Analyze the degrees of the numerator and denominator
The numerator is $-5\sqrt{x}+x - 1=x - 5x^{\frac{1}{2}}-1$ (dominant term is $x$) and the denominator is $-4x + 2$ (dominant term is $-4x$).
## Step2: Evaluate the limit
$\lim_{x
ightarrow\infty}\frac{-5\sqrt{x}+x - 1}{-4x + 2}=\lim_{x
ightarrow\infty}\frac{x(1-\frac{5}{\sqrt{x}}-\frac{1}{x})}{x(-4+\frac{2}{x})}=\frac{1-0 - 0}{-4+0}=-\frac{1}{4}$.

# Answer:
$-\frac{1}{4}$

### 59)
# Explanation:
## Step1: Divide numerator and denominator by highest - power of x in denominator
The highest - power of $x$ in the denominator is $x^{-2}$. Divide both numerator and denominator of $\frac{3x^{-1}-2x^{-3}}{3x^{-2}+x^{-5}}$ by $x^{-2}$. We get $\lim_{x
ightarrow\infty}\frac{3x - 2x^{-1}}{3+x^{-3}}$.
## Step2: Evaluate the limit
As $x
ightarrow\infty$, $x^{-1}
ightarrow0$ and $x^{-3}
ightarrow0$. So, $\lim_{x
ightarrow\infty}\frac{3x - 2x^{-1}}{3+x^{-3}}=\infty$.

# Answer:
$\infty$

### 60)
# Explanation:
## Step1: Divide numerator and denominator by highest - power of x in denominator
The highest - power of $x$ in the denominator is $x$. Divide both numerator and denominator of $\frac{\sqrt[3]{x}+5x - 5}{3x+x^{\frac{2}{3}}-4}$ by $x$. We get $\lim_{x
ightarrow-\infty}\frac{x^{-\frac{2}{3}}+5-\frac{5}{x}}{3 + x^{-\frac{1}{3}}-\frac{4}{x}}$.
## Step2: Evaluate the limit
As $x
ightarrow-\infty$, $x^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{x^{2}}}>0$, $x^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{x}}<0$, $\frac{5}{x}
ightarrow0$ and $\frac{4}{x}
ightarrow0$. So, $\lim_{x
ightarrow-\infty}\frac{x^{-\frac{2}{3}}+5-\frac{5}{x}}{3 + x^{-\frac{1}{3}}-\frac{4}{x}}=\frac{0 + 5-0}{3+0 - 0}=\frac{5}{3}$.

# Answer:
$\frac{5}{3}$

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QUESTION IMAGE

Question

Explanation:

55)

Step1: Divide numerator and denominator by highest - power of x in denominator

Step2: Evaluate the limit

Step1: Use limit properties

Step2: Substitute the limit values

Step1: Apply the Squeeze Theorem

Step2: Evaluate the limits of the bounding functions

Answer:

56)