If `lim_(ntooo) (n.3^(n))/(n(x-2)^(n)+n.3^(n+1)-3^(n))=1/3`, then the range of x is (where `n in N`)

Options

A. 1
B. `(2//3)^(1//2)`
C. `(3//2)^(1//2)`
D. `e^(1//2)`

Explanation

To solve the limit problem given, we start with the expression:

\[ \lim_{n \to \infty} \frac{n \cdot 3^n}{n(x-2)^n + n \cdot 3^{n+1} - 3^n} = \frac{1}{3} \]

### Step 1: Analyze the Limit
As \( n \) approaches infinity, both the numerator and denominator approach infinity. We can rewrite the limit by dividing both the numerator and denominator by \( 3^n \):

\[ \lim_{n \to \infty} \frac{n \cdot 3^n}{n(x-2)^n + n \cdot 3^{n+1} - 3^n} = \lim_{n \to \infty} \frac{n}{n \cdot \frac{(x-2)^n}{3^n} + n \cdot 3 - \frac{3^n}{3^n}} \]

### Step 2: Simplify the Expression
This simplifies to:

\[ \lim_{n \to \infty} \frac{n}{n \left( \frac{(x-2)^n}{3^n} + 3 - \frac{1}{n} \right)} \]

Cancelling \( n \) from the numerator and denominator gives:

\[ \lim_{n \to \infty} \frac{1}{\frac{(x-2)^n}{3^n} + 3 - \frac{1}{n}} \]

### Step 3: Analyze the Behavior of the Terms
As \( n \to \infty \), we need to analyze the term \( \frac{(x-2)^n}{3^n} \).

- If \( |x-2| < 3 \), then \( \frac{(x-2)^n}{3^n} \to 0 \).
- If \( |x-2| > 3 \), then \( \frac{(x-2)^n}{3^n} \to \infty \).
- If \( |x-2| = 3 \), then \( \frac{(x-2)^n}{3^n} \) approaches a constant.

### Step 4: Set Up the Equation
Since we want the limit to equal \( \frac{1}{3} \), we set up the equation:

\[ \frac{1}{0 + 3 - 0} = \frac{1}{3} \]

This implies that \( \frac{(x-2)^n}{3^n} \) must approach \( 0 \). Therefore, we require:

\[ |x-2| < 3 \]

### Step 5: Solve the Inequality
This inequality can be expressed as:

\[ -3 < x - 2 < 3 \]

Adding \( 2 \) to all parts gives:

\[ -1 < x < 5 \]

### Conclusion
Thus, the range of \( x \) is:

\[ (2 - 3, 2 + 3) = (-1, 5) \]

### Final Answer
The range of \( x \) is \( (-1, 5) \).