Solved In Exercises 11 and Finding Limits at Infinity 12, | Chegg.com - Free Printable

Problem Analysis:

The task involves finding limits at infinity for various functions. We will solve each part step by step, focusing on the behavior of the functions as \( x \to \infty \).

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Exercise 11:

Given \( f(x) = 5x^3 - 3 \), we need to find \( \lim_{x \to \infty} h(x) \) for the following cases:

#### (a) \( h(x) = \frac{f(x)}{x^2} \)
\[
h(x) = \frac{f(x)}{x^2} = \frac{5x^3 - 3}{x^2}
\]
Divide both the numerator and the denominator by \( x^2 \):
\[
h(x) = \frac{5x^3/x^2 - 3/x^2}{x^2/x^2} = \frac{5x - \frac{3}{x^2}}{1}
\]
As \( x \to \infty \):
- \( 5x \to \infty \)
- \( \frac{3}{x^2} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( 5x - \frac{3}{x^2} \right) = \infty
\]

#### (b) \( h(x) = \frac{f(x)}{x^3} \)
\[
h(x) = \frac{f(x)}{x^3} = \frac{5x^3 - 3}{x^3}
\]
Divide both the numerator and the denominator by \( x^3 \):
\[
h(x) = \frac{5x^3/x^3 - 3/x^3}{x^3/x^3} = \frac{5 - \frac{3}{x^3}}{1}
\]
As \( x \to \infty \):
- \( \frac{3}{x^3} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( 5 - \frac{3}{x^3} \right) = 5
\]

#### (c) \( h(x) = \frac{f(x)}{x^4} \)
\[
h(x) = \frac{f(x)}{x^4} = \frac{5x^3 - 3}{x^4}
\]
Divide both the numerator and the denominator by \( x^4 \):
\[
h(x) = \frac{5x^3/x^4 - 3/x^4}{x^4/x^4} = \frac{\frac{5}{x} - \frac{3}{x^4}}{1}
\]
As \( x \to \infty \):
- \( \frac{5}{x} \to 0 \)
- \( \frac{3}{x^4} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( \frac{5}{x} - \frac{3}{x^4} \right) = 0
\]

---

Exercise 12:

Given \( f(x) = -4x^2 + 2x - 5 \), we need to find \( \lim_{x \to \infty} h(x) \) for the following cases:

#### (a) \( h(x) = \frac{f(x)}{x} \)
\[
h(x) = \frac{f(x)}{x} = \frac{-4x^2 + 2x - 5}{x}
\]
Divide both the numerator and the denominator by \( x \):
\[
h(x) = \frac{-4x^2/x + 2x/x - 5/x}{x/x} = \frac{-4x + 2 - \frac{5}{x}}{1}
\]
As \( x \to \infty \):
- \( -4x \to -\infty \)
- \( 2 \to 2 \)
- \( \frac{5}{x} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( -4x + 2 - \frac{5}{x} \right) = -\infty
\]

#### (b) \( h(x) = \frac{f(x)}{x^2} \)
\[
h(x) = \frac{f(x)}{x^2} = \frac{-4x^2 + 2x - 5}{x^2}
\]
Divide both the numerator and the denominator by \( x^2 \):
\[
h(x) = \frac{-4x^2/x^2 + 2x/x^2 - 5/x^2}{x^2/x^2} = \frac{-4 + \frac{2}{x} - \frac{5}{x^2}}{1}
\]
As \( x \to \infty \):
- \( \frac{2}{x} \to 0 \)
- \( \frac{5}{x^2} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( -4 + \frac{2}{x} - \frac{5}{x^2} \right) = -4
\]

#### (c) \( h(x) = \frac{f(x)}{x^3} \)
\[
h(x) = \frac{f(x)}{x^3} = \frac{-4x^2 + 2x - 5}{x^3}
\]
Divide both the numerator and the denominator by \( x^3 \):
\[
h(x) = \frac{-4x^2/x^3 + 2x/x^3 - 5/x^3}{x^3/x^3} = \frac{-\frac{4}{x} + \frac{2}{x^2} - \frac{5}{x^3}}{1}
\]
As \( x \to \infty \):
- \( \frac{4}{x} \to 0 \)
- \( \frac{2}{x^2} \to 0 \)
- \( \frac{5}{x^3} \to 0 \)

Thus:
\[
\lim_{x \to \infty} h(x) = \lim_{x \to \infty} \left( -\frac{4}{x} + \frac{2}{x^2} - \frac{5}{x^3} \right) = 0
\]

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Exercise 13:

Find each limit if it exists.

#### (a) \( \lim_{x \to \infty} \frac{x^2 + 2}{x^3 - 1} \)
Divide both the numerator and the denominator by \( x^3 \):
\[
\frac{x^2 + 2}{x^3 - 1} = \frac{\frac{x^2}{x^3} + \frac{2}{x^3}}{\frac{x^3}{x^3} - \frac{1}{x^3}} = \frac{\frac{1}{x} + \frac{2}{x^3}}{1 - \frac{1}{x^3}}
\]
As \( x \to \infty \):
- \( \frac{1}{x} \to 0 \)
- \( \frac{2}{x^3} \to 0 \)
- \( \frac{1}{x^3} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{x^2 + 2}{x^3 - 1} = \frac{0 + 0}{1 - 0} = 0
\]

#### (b) \( \lim_{x \to \infty} \frac{x^2 + 2}{x^2 - 1} \)
Divide both the numerator and the denominator by \( x^2 \):
\[
\frac{x^2 + 2}{x^2 - 1} = \frac{\frac{x^2}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \frac{1 + \frac{2}{x^2}}{1 - \frac{1}{x^2}}
\]
As \( x \to \infty \):
- \( \frac{2}{x^2} \to 0 \)
- \( \frac{1}{x^2} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{x^2 + 2}{x^2 - 1} = \frac{1 + 0}{1 - 0} = 1
\]

#### (c) \( \lim_{x \to \infty} \frac{x^2 + 2}{x - 1} \)
Divide both the numerator and the denominator by \( x \):
\[
\frac{x^2 + 2}{x - 1} = \frac{\frac{x^2}{x} + \frac{2}{x}}{\frac{x}{x} - \frac{1}{x}} = \frac{x + \frac{2}{x}}{1 - \frac{1}{x}}
\]
As \( x \to \infty \):
- \( \frac{2}{x} \to 0 \)
- \( \frac{1}{x} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{x^2 + 2}{x - 1} = \frac{x + 0}{1 - 0} = \infty
\]

---

Exercise 14:

Find each limit if it exists.

#### (a) \( \lim_{x \to \infty} \frac{3 - 2x}{3x^3 - 1} \)
Divide both the numerator and the denominator by \( x^3 \):
\[
\frac{3 - 2x}{3x^3 - 1} = \frac{\frac{3}{x^3} - \frac{2x}{x^3}}{\frac{3x^3}{x^3} - \frac{1}{x^3}} = \frac{\frac{3}{x^3} - \frac{2}{x^2}}{3 - \frac{1}{x^3}}
\]
As \( x \to \infty \):
- \( \frac{3}{x^3} \to 0 \)
- \( \frac{2}{x^2} \to 0 \)
- \( \frac{1}{x^3} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{3 - 2x}{3x^3 - 1} = \frac{0 - 0}{3 - 0} = 0
\]

#### (b) \( \lim_{x \to \infty} \frac{3 - 2x}{3x - 1} \)
Divide both the numerator and the denominator by \( x \):
\[
\frac{3 - 2x}{3x - 1} = \frac{\frac{3}{x} - \frac{2x}{x}}{\frac{3x}{x} - \frac{1}{x}} = \frac{\frac{3}{x} - 2}{3 - \frac{1}{x}}
\]
As \( x \to \infty \):
- \( \frac{3}{x} \to 0 \)
- \( \frac{1}{x} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{3 - 2x}{3x - 1} = \frac{0 - 2}{3 - 0} = -\frac{2}{3}
\]

#### (c) \( \lim_{x \to \infty} \frac{3 - 2x^2}{3x - 1} \)
Divide both the numerator and the denominator by \( x \):
\[
\frac{3 - 2x^2}{3x - 1} = \frac{\frac{3}{x} - \frac{2x^2}{x}}{\frac{3x}{x} - \frac{1}{x}} = \frac{\frac{3}{x} - 2x}{3 - \frac{1}{x}}
\]
As \( x \to \infty \):
- \( \frac{3}{x} \to 0 \)
- \( 2x \to \infty \)
- \( \frac{1}{x} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{3 - 2x^2}{3x - 1} = \frac{0 - \infty}{3 - 0} = -\infty
\]

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Exercise 15:

Find each limit if it exists.

#### (a) \( \lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^{3/2} - 4} \)
Divide both the numerator and the denominator by \( x^{3/2} \):
\[
\frac{5 - 2x^{3/2}}{3x^{3/2} - 4} = \frac{\frac{5}{x^{3/2}} - \frac{2x^{3/2}}{x^{3/2}}}{\frac{3x^{3/2}}{x^{3/2}} - \frac{4}{x^{3/2}}} = \frac{\frac{5}{x^{3/2}} - 2}{3 - \frac{4}{x^{3/2}}}
\]
As \( x \to \infty \):
- \( \frac{5}{x^{3/2}} \to 0 \)
- \( \frac{4}{x^{3/2}} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^{3/2} - 4} = \frac{0 - 2}{3 - 0} = -\frac{2}{3}
\]

#### (b) \( \lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^{3/2} - 4x} \)
Divide both the numerator and the denominator by \( x^{3/2} \):
\[
\frac{5 - 2x^{3/2}}{3x^{3/2} - 4x} = \frac{\frac{5}{x^{3/2}} - \frac{2x^{3/2}}{x^{3/2}}}{\frac{3x^{3/2}}{x^{3/2}} - \frac{4x}{x^{3/2}}} = \frac{\frac{5}{x^{3/2}} - 2}{3 - \frac{4}{x^{1/2}}}
\]
As \( x \to \infty \):
- \( \frac{5}{x^{3/2}} \to 0 \)
- \( \frac{4}{x^{1/2}} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x^{3/2} - 4x} = \frac{0 - 2}{3 - 0} = -\frac{2}{3}
\]

#### (c) \( \lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x - 4} \)
Divide both the numerator and the denominator by \( x \):
\[
\frac{5 - 2x^{3/2}}{3x - 4} = \frac{\frac{5}{x} - \frac{2x^{3/2}}{x}}{\frac{3x}{x} - \frac{4}{x}} = \frac{\frac{5}{x} - 2x^{1/2}}{3 - \frac{4}{x}}
\]
As \( x \to \infty \):
- \( \frac{5}{x} \to 0 \)
- \( 2x^{1/2} \to \infty \)
- \( \frac{4}{x} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5 - 2x^{3/2}}{3x - 4} = \frac{0 - \infty}{3 - 0} = -\infty
\]

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Exercise 16:

Find each limit if it exists.

#### (a) \( \lim_{x \to \infty} \frac{5x^{3/2}}{4x^2 + 1} \)
Divide both the numerator and the denominator by \( x^2 \):
\[
\frac{5x^{3/2}}{4x^2 + 1} = \frac{\frac{5x^{3/2}}{x^2}}{\frac{4x^2}{x^2} + \frac{1}{x^2}} = \frac{\frac{5}{x^{1/2}}}{4 + \frac{1}{x^2}}
\]
As \( x \to \infty \):
- \( \frac{5}{x^{1/2}} \to 0 \)
- \( \frac{1}{x^2} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5x^{3/2}}{4x^2 + 1} = \frac{0}{4 + 0} = 0
\]

#### (b) \( \lim_{x \to \infty} \frac{5x^{3/2}}{4x^{3/2} + 1} \)
Divide both the numerator and the denominator by \( x^{3/2} \):
\[
\frac{5x^{3/2}}{4x^{3/2} + 1} = \frac{\frac{5x^{3/2}}{x^{3/2}}}{\frac{4x^{3/2}}{x^{3/2}} + \frac{1}{x^{3/2}}} = \frac{5}{4 + \frac{1}{x^{3/2}}}
\]
As \( x \to \infty \):
- \( \frac{1}{x^{3/2}} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5x^{3/2}}{4x^{3/2} + 1} = \frac{5}{4 + 0} = \frac{5}{4}
\]

#### (c) \( \lim_{x \to \infty} \frac{5x^{3/2}}{4\sqrt{x} + 1} \)
Divide both the numerator and the denominator by \( \sqrt{x} \):
\[
\frac{5x^{3/2}}{4\sqrt{x} + 1} = \frac{\frac{5x^{3/2}}{\sqrt{x}}}{\frac{4\sqrt{x}}{\sqrt{x}} + \frac{1}{\sqrt{x}}} = \frac{5x}{4 + \frac{1}{\sqrt{x}}}
\]
As \( x \to \infty \):
- \( \frac{1}{\sqrt{x}} \to 0 \)

Thus:
\[
\lim_{x \to \infty} \frac{5x^{3/2}}{4\sqrt{x} + 1} = \frac{5x}{4 + 0} = \infty
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
&\text{Exercise 11:} \\
&(a) \infty, \quad (b) 5, \quad (c) 0 \\
&\text{Exercise 12:} \\
&(a) -\infty, \quad (b) -4, \quad (c) 0 \\
&\text{Exercise 13:} \\
&(a) 0, \quad (b) 1, \quad (c) \infty \\
&\text{Exercise 14:} \\
&(a) 0, \quad (b) -\frac{2}{3}, \quad (c) -\infty \\
&\text{Exercise 15:} \\
&(a) -\frac{2}{3}, \quad (b) -\frac{2}{3}, \quad (c) -\infty \\
&\text{Exercise 16:} \\
&(a) 0, \quad (b) \frac{5}{4}, \quad (c) \infty
\end{aligned}
}
\]