Let's solve the given problems one by one.

Problem (a):

$\lim_{x \to \infty} \left( \ln(8x^4 + 7) - \ln(2x^3 - 3) \right)$

Using logarithmic properties, we can combine the logarithms: $\lim_{x \to \infty} \ln\left( \frac{8x^4 + 7}{2x^3 - 3} \right)$

Now, let's simplify the expression inside the logarithm: $\frac{8x^4 + 7}{2x^3 - 3} \approx \frac{8x^4}{2x^3} = 4x \quad \text{as} \quad x \to \infty$

So, the limit becomes: $\lim_{x \to \infty} \ln(4x)$

As $x \to \infty$, $\ln(4x) \to \infty$, hence: $\lim_{x \to \infty} \ln(4x) = \infty$

Thus, the limit for part (a) is infinity.

Problem (b):

$\lim_{x \to \infty} \frac{e^x - e^{-x}}{6e^{-x} + 5e^x}$

First, divide both the numerator and the denominator by $e^x$ to simplify: $\frac{e^x - e^{-x}}{6e^{-x} + 5e^x} = \frac{1 - e^{-2x}}{\frac{6e^{-x}}{e^x} + 5} = \frac{1 - e^{-2x}}{6e^{-2x} + 5}$

As $x \to \infty$, $e^{-2x} \to 0$. Therefore, the expression simplifies to: $\frac{1 - 0}{0 + 5} = \frac{1}{5}$

Thus, the limit for part (b) is $\frac{1}{5}$.

Would you like more detailed explanations or have any other questions?

Here are 5 related questions you can try:

1. What happens if the powers in the logarithmic expression of part (a) were different?
2. How does the limit change if the coefficients in part (b) are altered?
3. How do limits involving exponential functions behave as $x \to -\infty$?
4. Can you compute the limit $\lim_{x \to \infty} \frac{\ln(x)}{x^2}$?
5. How do we approach limits involving trigonometric functions at infinity?

Tip: Always check if simplifying the expression (like factoring or using properties of logarithms) can help solve limits more easily!