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.