To solve $\lim_{{x \to \infty}} \left( \frac{x+1}{x-2} \right)^{2x-1}$, let's proceed step-by-step.

1. Rewrite the base expression: $\frac{x+1}{x-2} = \frac{x(1 + \frac{1}{x})}{x(1 - \frac{2}{x})} = \frac{1 + \frac{1}{x}}{1 - \frac{2}{x}}$

   As $x \to \infty$, both $\frac{1}{x} \to 0$ and $\frac{2}{x} \to 0$. So, the expression inside the limit approaches: $\frac{1 + 0}{1 - 0} = 1$

2. Approximate the base using a binomial expansion: Since $\frac{1 + \frac{1}{x}}{1 - \frac{2}{x}} \approx 1 + \frac{1}{x} + \frac{2}{x} = 1 + \frac{3}{x}$, we can approximate it by: $\frac{x+1}{x-2} \approx 1 + \frac{3}{x}$

3. Logarithmic approximation: For expressions close to 1, we use the approximation $\ln(1 + y) \approx y$ when $y \approx 0$.

   $\ln\left( \frac{x+1}{x-2} \right) \approx \ln\left( 1 + \frac{3}{x} \right) \approx \frac{3}{x}$

4. Apply logarithmic transformation: Take the natural logarithm of the limit expression: $\ln \left[ \left( \frac{x+1}{x-2} \right)^{2x-1} \right] = (2x-1) \ln \left( \frac{x+1}{x-2} \right)$ Now substitute the approximation for the logarithm: $(2x-1) \cdot \frac{3}{x} = \frac{3(2x-1)}{x} = 6 - \frac{3}{x}$

   As $x \to \infty$, $- \frac{3}{x} \to 0$, so the expression approaches: $6$

5. Exponential transformation: Therefore, the original limit becomes: $\lim_{{x \to \infty}} \left( \frac{x+1}{x-2} \right)^{2x-1} = e^{6}$

The final answer is:

$\lim_{{x \to \infty}} \left( \frac{x+1}{x-2} \right)^{2x-1} = e^{6}$

Would you like more details on any step, or do you have any specific questions?

Here are 5 relative questions to extend your understanding:

1. How do we apply L'Hospital's Rule to limits involving indeterminate forms?
2. What is the Taylor series expansion, and how does it relate to the approximation used here?
3. How can we find the limit of a general expression raised to a power as it approaches infinity?
4. What are common indeterminate forms and methods to solve them?
5. How can logarithmic differentiation help in solving complex limits?

Tip: For limits approaching infinity, transforming the problem using logarithms can often simplify finding the solution, especially for expressions raised to a power.