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Let's go over the two limit problems step by step.

### Problem d:
\[
\lim_{{x \to \infty}} \frac{\sqrt{16x^2 + 3x - 7}}{4x - 4}
\]

#### Steps to Solve:
1. **Analyze the numerator**: The highest degree term inside the square root is \(16x^2\). We can factor out \(x^2\) from the expression under the square root:
   \[
   \sqrt{16x^2 + 3x - 7} = \sqrt{x^2(16 + \frac{3}{x} - \frac{7}{x^2})} = |x|\sqrt{16 + \frac{3}{x} - \frac{7}{x^2}}.
   \]
   Since \(x \to \infty\), \(|x| = x\), and the expression simplifies to:
   \[
   x\sqrt{16 + \frac{3}{x} - \frac{7}{x^2}}.
   \]

2. **Simplify the denominator**: The denominator is \(4x - 4\), which factors as:
   \[
   4x - 4 = 4(x - 1).
   \]

3. **Form the ratio**: Now the limit becomes:
   \[
   \lim_{{x \to \infty}} \frac{x\sqrt{16 + \frac{3}{x} - \frac{7}{x^2}}}{4(x - 1)}.
   \]

4. **Take the limit**: As \(x \to \infty\), the terms \(\frac{3}{x}\) and \(\frac{7}{x^2}\) tend to zero, so:
   \[
   \sqrt{16 + \frac{3}{x} - \frac{7}{x^2}} \to \sqrt{16} = 4.
   \]
   Thus, the limit simplifies to:
   \[
   \lim_{{x \to \infty}} \frac{x \cdot 4}{4(x - 1)} = \lim_{{x \to \infty}} \frac{4x}{4x - 4}.
   \]
   Factoring out \(x\) in the denominator gives:
   \[
   \lim_{{x \to \infty}} \frac{4x}{4(x - 1)} = \lim_{{x \to \infty}} \frac{4x}{4x(1 - \frac{1}{x})} = \lim_{{x \to \infty}} \frac{4}{4(1 - \frac{1}{x})} = \frac{4}{4} = 1.
   \]

So, the answer to problem **d** is:
\[
\boxed{1}.
\]

---

### Problem e:
\[
\lim_{{x \to \infty}} \frac{(2x + 3)^3 + 5x^2}{(3x - 1)(x + 2)^2}
\]

#### Steps to Solve:
1. **Expand and simplify the numerator**: The term \((2x + 3)^3\) can be expanded, but we only care about the leading term as \(x \to \infty\). The leading term of \((2x + 3)^3\) is \(8x^3\), so the numerator behaves as:
   \[
   (2x + 3)^3 + 5x^2 \approx 8x^3 + 5x^2.
   \]

2. **Expand the denominator**: We expand \((x + 2)^2\), which gives:
   \[
   (x + 2)^2 = x^2 + 4x + 4.
   \]
   So, the denominator becomes:
   \[
   (3x - 1)(x^2 + 4x + 4) \approx 3x(x^2 + 4x + 4) = 3x^3 + 12x^2 + 12x.
   \]

3. **Form the ratio**: Now, the limit becomes:
   \[
   \lim_{{x \to \infty}} \frac{8x^3 + 5x^2}{3x^3 + 12x^2 + 12x}.
   \]

4. **Divide by the highest power of \(x\)**: To simplify, divide both the numerator and denominator by \(x^3\):
   \[
   \lim_{{x \to \infty}} \frac{8 + \frac{5}{x}}{3 + \frac{12}{x} + \frac{12}{x^2}}.
   \]

5. **Take the limit**: As \(x \to \infty\), all terms with \(x\) in the denominator tend to zero, so:
   \[
   \frac{8 + 0}{3 + 0 + 0} = \frac{8}{3}.
   \]

So, the answer to problem **e** is:
\[
\boxed{\frac{8}{3}}.
\]

---

Do you have any questions or want further clarification on these problems? Here are some related questions for further practice:

1. What happens to limits involving logarithms as \(x \to \infty\)?
2. How do you apply L'Hopital's rule to indeterminate forms like \(\frac{0}{0}\)?
3. How would you solve limits at infinity involving trigonometric functions?
4. Can limits at infinity sometimes be undefined?
5. How do polynomial degrees affect limits at infinity?

**Tip**: When dealing with limits at infinity, always focus on the highest degree terms, as they dominate the behavior of the function.

Solve the following limits as x approaches infinity: 
1. lim(x→∞) [√(16x^2 + 3x - 7) / (4x - 4)] 
2. lim(x→∞) [(2x + 3)^3 + 5x^2] / [(3x - 1)(x + 2)^2]

Step-by-step guide to solving limits at infinity for polynomial and rational functions. Learn the techniques for finding the behavior of complex expressions as x approaches infinity. Suitable for high school students.

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