We need to evaluate the limit:

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

Step 1: Use Log Approximation

For small $y$, we use the approximation:

$\ln(1 + y) \approx y \quad \text{for } y \approx 0$

Setting $y = -\frac{5}{x} + \frac{6}{x^2}$, we assume $y$ is small for large $x$, so we approximate:

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

Step 2: Multiply by $x$

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

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

$= -5 + \frac{6}{x}$

Step 3: Evaluate the Limit

$\lim_{x \to \infty} \left( -5 + \frac{6}{x} \right) = -5 + 0 = -5$

Conclusion

Thus, the given limit evaluates to:

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

Would you like a more detailed breakdown or an alternative approach?

Here are five related problems for practice:

1. $\lim_{x \to \infty} x \ln(1 + \frac{a}{x})$ where $a$ is a constant.
2. $\lim_{x \to \infty} x \ln(1 - \frac{b}{x} + \frac{c}{x^2})$ for arbitrary constants $b, c$.
3. $\lim_{x \to \infty} x \ln(1 + \frac{1}{x^2})$.
4. $\lim_{x \to \infty} x^2 \ln(1 - \frac{1}{x})$.
5. $\lim_{x \to \infty} x \ln(1 - \frac{k}{x^3})$ where $k$ is a constant.

Tip:

For limits involving logarithms and small perturbations, always consider the first-order approximation:
$\ln(1 + y) \approx y \text{ for small } y$
This simplifies calculations significantly! 🚀