We are tasked with finding the limit of $f(x) = \frac{2^x}{3^x - 1}$ as $x \to \infty$.

Step-by-Step Solution:

1. Rewrite the function: $f(x) = \frac{2^x}{3^x - 1}$ As $x \to \infty$, both $2^x$ and $3^x$ grow exponentially, but $3^x$ grows faster because the base (3) is larger than 2.

2. Factor the denominator to compare terms: Factor out $3^x$ from the denominator: $f(x) = \frac{2^x}{3^x(1 - \frac{1}{3^x})}$

3. Simplify the expression: $f(x) = \frac{\left( \frac{2^x}{3^x} \right)}{\left( 1 - \frac{1}{3^x} \right)} = \frac{\left( \left( \frac{2}{3} \right)^x \right)}{\left( 1 - \frac{1}{3^x} \right)}$

4. Analyze the behavior as $x \to \infty$:

   • As $x \to \infty$, $\left( \frac{2}{3} \right)^x \to 0$ because $\frac{2}{3} < 1$.
   • Also, $\frac{1}{3^x} \to 0$, so the denominator approaches 1.

5. Take the limit: $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\left( \frac{2}{3} \right)^x}{1 - 0} = 0$

Conclusion:

$\lim_{x \to \infty} \frac{2^x}{3^x - 1} = 0$

Would you like further details or have any questions? Here are 5 related problems you might find interesting:

1. Find $\lim_{x \to \infty} \frac{5^x}{6^x + 2}$.
2. What is $\lim_{x \to \infty} \frac{2x}{3x - 1}$?
3. Compute $\lim_{x \to \infty} \frac{e^x}{e^{2x} - 1}$.
4. Evaluate $\lim_{x \to \infty} \frac{3^x + 1}{5^x}$.
5. Determine $\lim_{x \to \infty} \frac{\ln(x)}{x}$.

Tip: When comparing exponential functions, the one with the larger base dominates as $x \to \infty$.