To evaluate the given limit:

$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{3x},$

Step 1: Recognizing the exponential form

The expression $\left(1 - \frac{2}{x}\right)^{3x}$ suggests using the exponential limit property: $\lim_{x \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k.$

Here, rewrite the base $1 - \frac{2}{x}$ as: $1 - \frac{2}{x} = 1 + \left(-\frac{2}{x}\right).$

Step 2: Applying logarithms

Let $y = \left(1 - \frac{2}{x}\right)^{3x}$. Taking the natural logarithm: $\ln y = 3x \ln\left(1 - \frac{2}{x}\right).$

Step 3: Using the approximation $\ln(1 + u) \approx u$ for small $u$

When $x \to \infty$, $-\frac{2}{x} \to 0$. Thus, expand $\ln(1 - \frac{2}{x})$ as: $\ln\left(1 - \frac{2}{x}\right) \approx -\frac{2}{x}.$

Substitute this into the expression for $\ln y$: $\ln y \approx 3x \cdot \left(-\frac{2}{x}\right).$

Simplify: $\ln y \approx -6.$

Step 4: Exponentiating to find $y$

Since $\ln y \approx -6$, we exponentiate: $y \approx e^{-6}.$

Final Answer:

$\lim_{x \to \infty} \left(1 - \frac{2}{x}\right)^{3x} = e^{-6}.$

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Related Questions:

1. What is the derivation of the exponential limit property $(1 + k/n)^n \to e^k$?
2. How do Taylor expansions justify the approximation $\ln(1 + u) \approx u$?
3. What are the general steps for solving limits involving exponential growth?
4. How does this limit compare to the form $\lim_{x \to \infty} (1 + k/x)^x$?
5. Can this problem be solved without logarithmic substitution?

Tip: Recognizing the exponential structure in limits simplifies calculations and avoids unnecessary computations.