`If("lim")_(xvecoo)(n .3^n)/(n(x-2)^n+n .3^(n+1)-3^n)=1/3, t h e n t h e` range of `x` is (where `n in N)dot` (a) (2,5) (b) (1,5) (c) `-5,5)` (d) `(-oo,oo)`

Options

A. 3
B. 4
C. 5
D. infinite

Explanation

To solve the limit problem given in the question, we need to analyze the expression step by step. The limit we need to evaluate is:

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

### Step 1: Simplify the Expression
First, we can rewrite the limit expression:

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

### Step 2: Factor Out \(3^n\)
Next, we can factor \(3^n\) out of the denominator:

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

This simplifies to:

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

### Step 3: Analyze the Limit
Now we need to analyze the behavior of \(\frac{(x-2)^n}{3^n}\) as \(n\) approaches infinity.

- If \(|x-2| < 3\), then \(\frac{(x-2)^n}{3^n} \to 0\).
- If \(|x-2| = 3\), then \(\frac{(x-2)^n}{3^n} \to 1\).
- If \(|x-2| > 3\), then \(\frac{(x-2)^n}{3^n} \to \infty\).

### Step 4: Evaluate the Limit Based on Cases
1. **Case 1**: If \(|x-2| < 3\):
- The limit becomes:
\[ \lim_{n \to \infty} \frac{n}{0 + 3n - 1} = \lim_{n \to \infty} \frac{n}{3n} = \frac{1}{3} \]

2. **Case 2**: If \(|x-2| = 3\):
- The limit becomes:
\[ \lim_{n \to \infty} \frac{n}{n + 3n - 1} = \lim_{n \to \infty} \frac{n}{4n - 1} = \frac{1}{4} \]

3. **Case 3**: If \(|x-2| > 3\):
- The limit diverges to infinity.

### Step 5: Determine the Range of \(x\)
For the limit to equal \(\frac{1}{3}\), we require:

\[ |x-2| < 3 \implies -3 < x - 2 < 3 \implies -1 < x < 5 \]

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

\[ (-1, 5) \]

### Final Answer
The correct option that matches our derived range is:

(b) \((1, 5)\)