Let's go through your given limit and then explore alternative ways to approach it.

The limit you're working with is:

$\lim_{x \to \infty} x \ln\left( 1 - \frac{5}{x} + \frac{6}{x^2} \right)$

Step 1: Simplifying the expression

As $x \to \infty$, both $\frac{5}{x}$ and $\frac{6}{x^2}$ approach 0, so the argument inside the logarithm approaches 1. However, since $x$ is also tending to infinity, the expression is an indeterminate form of the type $\infty \times 0$. This suggests that we can look for a way to simplify it further using series expansions or approximations.

Step 2: Expanding the logarithmic term

We can use a series expansion for $\ln(1 + u)$ around $u = 0$, which is:

$\ln(1 + u) \approx u \quad \text{for small} \ u$

In this case, the argument of the logarithm is $1 - \frac{5}{x} + \frac{6}{x^2}$, so we can rewrite it as:

$\ln\left( 1 - \frac{5}{x} + \frac{6}{x^2} \right) \approx -\frac{5}{x} + \frac{6}{x^2}$

Step 3: Substituting the approximation

Now substitute the approximation of the logarithm back into the original limit expression:

$x \left( -\frac{5}{x} + \frac{6}{x^2} \right) = -5 + \frac{6}{x}$

Step 4: Taking the limit

As $x \to \infty$, the term $\frac{6}{x}$ approaches 0. So, we are left with:

$\lim_{x \to \infty} \left( -5 + \frac{6}{x} \right) = -5$

Thus, the limit is:

$\boxed{-5}$

Other Ways to Approach:

1. Taylor Expansion: Another way to approach this is by expanding the argument of the logarithm using a Taylor series for $\ln(1 + u)$ and then simplifying the product with $x$.
2. L'Hopital's Rule: You could apply L'Hopital's Rule to the indeterminate form $\infty \times 0$ after transforming the expression into a ratio.

Would you like more details on any of these methods?

Here are some related questions you might explore:

1. How does the Taylor series for $\ln(1 + u)$ work in more general cases?
2. What happens if you use L'Hopital's Rule here? How do we transform the problem into a fraction?
3. Can this method be generalized to other limits involving indeterminate forms of type $\infty \times 0$?
4. What are the implications of higher-order terms in the series expansion for this type of limit?
5. How do we handle limits involving logarithmic functions in more complex cases?

Tip: Always check for indeterminate forms like $\infty \times 0$ or $0/0$, as they are often signals for simplifications using series or L'Hopital's Rule.