What is the limit of x as x approaches infinity of f(x)=(2^x)/(3^x-1)

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 \( x \to \infty \).

What is the limit of x as x approaches infinity of f(x)=(2^x)/(3^x-1)?

Learn how to find the limit of the function (2^x)/(3^x - 1) as x approaches infinity. This detailed step-by-step solution covers the behavior of exponential functions and the limit comparison theorem. Suitable for students in Grades 10-12.

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