https://file.aiquickdraw.com/mathbot/file/1725976365737oy1dmqni.jpg

Let's go through the questions one by one, solving the limits provided in the image.

### 1. 
\[
\lim_{x \to \infty} \frac{3x^3 - 2x - 10}{4x - 2x^2 - 5x^3}
\]

To solve this limit, divide both the numerator and denominator by the highest power of \(x\), which is \(x^3\):

\[
\lim_{x \to \infty} \frac{\frac{3x^3}{x^3} - \frac{2x}{x^3} - \frac{10}{x^3}}{\frac{4x}{x^3} - \frac{2x^2}{x^3} - \frac{5x^3}{x^3}}
= \lim_{x \to \infty} \frac{3 - \frac{2}{x^2} - \frac{10}{x^3}}{\frac{4}{x^2} - \frac{2}{x} - 5}
\]

As \(x \to \infty\), all terms with \(x\) in the denominator tend to 0:

\[
\frac{3 - 0 - 0}{0 - 0 - 5} = \frac{3}{-5} = -\frac{3}{5}
\]

The correct answer is **A.** \(-\frac{3}{5}\).

---

### 2. 
\[
\lim_{x \to \infty} \frac{x^5 - 2x^4 + x^3 - 3x^2 + 2x - 7}{7 - 2x + 3x^2 - x^3 + 2x^4}
\]

Again, divide the numerator and denominator by the highest power of \(x\), which is \(x^5\):

\[
\lim_{x \to \infty} \frac{\frac{x^5}{x^5} - \frac{2x^4}{x^5} + \frac{x^3}{x^5} - \frac{3x^2}{x^5} + \frac{2x}{x^5} - \frac{7}{x^5}}{\frac{7}{x^5} - \frac{2x}{x^5} + \frac{3x^2}{x^5} - \frac{x^3}{x^5} + \frac{2x^4}{x^5}}
\]

This simplifies to:

\[
\lim_{x \to \infty} \frac{1 - \frac{2}{x} + \frac{1}{x^2} - \frac{3}{x^3} + \frac{2}{x^4} - \frac{7}{x^5}}{\frac{7}{x^5} - \frac{2}{x^4} + \frac{3}{x^3} - \frac{1}{x^2} + \frac{2}{x}}
\]

As \(x \to \infty\), the terms with powers of \(x\) in the denominator vanish, leaving:

\[
\frac{1}{0} \rightarrow \infty
\]

The correct answer is **E.** \(\infty\).

---

### 3. 
\[
\lim_{x \to \infty} \frac{(1 - 2x)^3}{(x - 1)(2x^2 + x + 1)}
\]

First, expand the numerator and denominator.

The numerator:

\[
(1 - 2x)^3 = 1 - 6x + 12x^2 - 8x^3
\]

The denominator:

\[
(x - 1)(2x^2 + x + 1) = x(2x^2 + x + 1) - 1(2x^2 + x + 1) = 2x^3 + x^2 + x - 2x^2 - x - 1 = 2x^3 - x^2
\]

Now, divide both numerator and denominator by \(x^3\):

\[
\lim_{x \to \infty} \frac{\frac{1}{x^3} - \frac{6}{x^2} + \frac{12}{x} - 8}{2 - \frac{1}{x}}
\]

As \(x \to \infty\), this becomes:

\[
\frac{-8}{2} = -4
\]

The correct answer is **B.** \(-4\).

---

### 4. 
\[
\lim_{x \to \infty} \frac{(3x - 2)^3}{(4x + 2)^3}
\]

First, expand both the numerator and the denominator:

The numerator:

\[
(3x - 2)^3 = 27x^3 - 54x^2 + 36x - 8
\]

The denominator:

\[
(4x + 2)^3 = 64x^3 + 96x^2 + 48x + 8
\]

Now, divide both numerator and denominator by \(x^3\):

\[
\lim_{x \to \infty} \frac{\frac{27x^3}{x^3} - \frac{54x^2}{x^3} + \frac{36x}{x^3} - \frac{8}{x^3}}{\frac{64x^3}{x^3} + \frac{96x^2}{x^3} + \frac{48x}{x^3} + \frac{8}{x^3}} = \lim_{x \to \infty} \frac{27 - \frac{54}{x} + \frac{36}{x^2} - \frac{8}{x^3}}{64 + \frac{96}{x} + \frac{48}{x^2} + \frac{8}{x^3}}
\]

As \(x \to \infty\), this becomes:

\[
\frac{27}{64}
\]

The correct answer is **B.** \(\frac{27}{64}\).

---

Would you like further details on any of these steps, or do you have any questions?

Here are 5 related questions you might find useful:
1. What are the key techniques for solving limits at infinity?
2. How do you handle indeterminate forms in limits?
3. What is the role of L'Hopital's Rule in solving limits involving infinity?
4. How do polynomials behave as \(x\) approaches infinity?
5. When should you apply factorization techniques in solving limits?

**Tip:** Always remember to divide by the highest power of \(x\) in rational functions when taking limits as \(x \to \infty\).

The image contains four mathematical limit problems involving polynomials and rational functions at infinity.

Solve these challenging limit problems involving rational functions and polynomials. This set of problems focuses on understanding the limits of functions as x approaches infinity, with detailed solutions covering simplifying expressions and applying dominant term analysis.

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