Let's analyze the given limit:

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

Step 1: Simplifying the expression inside the logarithm

As $x \to \infty$, the terms $\frac{5}{x}$ and $\frac{6}{x^2}$ become very small. Therefore, we can approximate:

$1 - \frac{5}{x} + \frac{6}{x^2} \approx 1 - \frac{5}{x}$

The term $\frac{6}{x^2}$ becomes negligible compared to $\frac{5}{x}$ for large $x$, so we approximate:

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

Step 2: Using the approximation for $\ln(1 + u)$ when $u$ is small

For small values of $u$, we can use the approximation:

$\ln(1 + u) \approx u \quad \text{when} \quad u \to 0$

Thus, for $u = -\frac{5}{x}$, we have:

$\ln\left(1 - \frac{5}{x}\right) \approx -\frac{5}{x}$

Step 3: Substituting this approximation back into the limit

Now, substitute $\ln\left(1 - \frac{5}{x}\right) \approx -\frac{5}{x}$ into the original limit:

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

Simplifying:

$= \lim_{x \to \infty} -5 = -5$

Final Answer:

The limit is $\boxed{-5}$.

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Would you like further details or have any other questions?

Here are some related questions you might find interesting:

1. What is the behavior of logarithmic functions as $x \to \infty$?
2. How can we use series expansions to approximate limits?
3. What happens to the limit if we have higher-order terms in the approximation?
4. How do we handle indeterminate forms like $0 \times \infty$ in limits?
5. Can this result be generalized to other similar expressions involving logarithms?

Tip: Always check whether approximations are valid for large $x$ before proceeding, especially in limits involving small terms like $\frac{1}{x}$.