This problem requires finding the limit of the given expression as $x \to \infty$. The expression is:

$\lim_{x \to \infty} \left(\frac{3x + 4}{3x - 4}\right)^{-5x}.$

Step-by-Step Solution:

Step 1: Simplify the fraction inside the limit

Divide the numerator and denominator of the fraction $\frac{3x + 4}{3x - 4}$ by $x$: $\frac{3x + 4}{3x - 4} = \frac{3 + \frac{4}{x}}{3 - \frac{4}{x}}.$

As $x \to \infty$, the terms $\frac{4}{x}$ approach $0$. Therefore: $\frac{3 + \frac{4}{x}}{3 - \frac{4}{x}} \to \frac{3 + 0}{3 - 0} = 1.$

However, since the expression is raised to the power $-5x$, we must carefully analyze its behavior for large $x$.

Step 2: Approximate $\frac{3x + 4}{3x - 4}$ for large $x$

For large $x$, use a first-order approximation: $\frac{3 + \frac{4}{x}}{3 - \frac{4}{x}} \approx 1 + \frac{\frac{4}{x} + \frac{4}{x}}{3} = 1 + \frac{8}{3x}.$

Thus, for large $x$, we can write: $\frac{3x + 4}{3x - 4} \approx 1 + \frac{8}{3x}.$

Step 3: Expand $(1 + u)^v$ using logarithms

Let $u = \frac{8}{3x}$. The original limit becomes: $\lim_{x \to \infty} \left(1 + \frac{8}{3x}\right)^{-5x}.$

Using the exponential-logarithmic property, rewrite it as: $\left(1 + \frac{8}{3x}\right)^{-5x} = \exp\left(-5x \ln\left(1 + \frac{8}{3x}\right)\right).$

For small $u$, $\ln(1 + u) \approx u$. Substituting $\ln\left(1 + \frac{8}{3x}\right) \approx \frac{8}{3x}$: $-5x \ln\left(1 + \frac{8}{3x}\right) \approx -5x \cdot \frac{8}{3x} = -\frac{40}{3}.$

Thus, the expression simplifies to: $\exp\left(-\frac{40}{3}\right).$

Final Answer:

$\lim_{x \to \infty} \left(\frac{3x + 4}{3x - 4}\right)^{-5x} = e^{-\frac{40}{3}}.$

Do you want a more detailed explanation of any step? Let me know!

Related Questions:

1. What is the behavior of exponential functions as their power grows?
2. How can logarithmic approximations help in solving limits?
3. Why is $\ln(1 + u) \approx u$ valid for small $u$?
4. What other cases involve limits with powers that grow large?
5. How would the solution differ if the power was $5x$ instead of $-5x$?

Tip:

Always check for approximations (like $\ln(1 + u) \approx u$) and validate their range of validity when solving limits involving exponential or logarithmic expressions!