To evaluate the given limit:

$\lim_{x \to \infty} \left( \frac{mx + n}{mx - n} \right)^{(m + n)x}$ where $m = 4$ and $n = 5$, let's proceed step by step:

Step 1: Simplify the Fraction

Divide both numerator and denominator by $x$ to factor out the dominant term:

$\frac{mx + n}{mx - n} = \frac{m + \frac{n}{x}}{m - \frac{n}{x}}.$

As $x \to \infty$, the terms $\frac{n}{x}$ approach zero:

$\frac{m + \frac{n}{x}}{m - \frac{n}{x}} \to \frac{m}{m} = 1.$

However, since this expression is raised to the power $(m + n)x$, we must consider how small deviations from 1 impact the limit.

Step 2: Approximate the Logarithm

To handle the exponentiation, take the natural logarithm of the expression:

$L = \ln \left( \left( \frac{mx + n}{mx - n} \right)^{(m + n)x} \right) = (m + n)x \cdot \ln \left( \frac{mx + n}{mx - n} \right).$

For small $y$, the approximation $\ln(1 + y) \approx y$ applies. Here:

\frac{mx + n}{mx - n} = 1 + \frac{2n}{mx - n} \quad \text{(for large x)}.

Thus, using the logarithmic approximation:

$\ln \left( \frac{mx + n}{mx - n} \right) \approx \frac{2n}{mx}.$

Step 3: Combine Terms

Substitute the approximation into $L$:

$L \approx (m + n)x \cdot \frac{2n}{mx} = \frac{2n(m + n)}{m}.$

For $m = 4$ and $n = 5$, we calculate:

$L = \frac{2(5)(4 + 5)}{4} = \frac{2 \cdot 5 \cdot 9}{4} = \frac{90}{4} = 22.5.$

Step 4: Final Result

Exponentiate $L$ to obtain the original limit:

$\lim_{x \to \infty} \left( \frac{4x + 5}{4x - 5} \right)^{9x} = e^{22.5}.$

Final Answer:

$e^{22.5}.$

Let me know if you'd like more details!

Related Questions:

1. How can we use logarithmic approximations for limits?
2. What is the behavior of powers near 1 as $x \to \infty$?
3. How do dominant terms affect limits involving exponents?
4. How does the exponential function relate to $e^x$?
5. Can L'Hôpital's Rule solve similar limits?

Tip: For limits involving powers, consider approximations like $\ln(1 + y) \approx y$ when $y \to 0$.