For `n epsilon N` let `x_(n)` be defined as `(1+1/n)^((n+x_(n)))=e` then `lim_(nto oo)(2x_(n))` equals…..

Answer

To solve the problem, we start with the given equation:

\[ (1 + \frac{1}{n})^{(n + x_n)} = e \]

We need to find the limit:

\[ \lim_{n \to \infty} (2x_n) \]

### Step 1: Rewrite the equation

We can rewrite the equation as:

\[ x_n = (1 + \frac{1}{n})^{(n + x_n)} = e \]

Taking the natural logarithm on both sides gives us:

\[ \ln((1 + \frac{1}{n})^{(n + x_n)}) = \ln(e) \]

This simplifies to:

\[ (n + x_n) \ln(1 + \frac{1}{n}) = 1 \]

### Step 2: Use the expansion of \(\ln(1 + x)\)

Using the Taylor series expansion for \(\ln(1 + x)\) around \(x = 0\):

\[ \ln(1 + \frac{1}{n}) \approx \frac{1}{n} - \frac{1}{2n^2} + O(\frac{1}{n^3}) \]

Substituting this into our equation gives:

\[ (n + x_n) \left(\frac{1}{n} - \frac{1}{2n^2} + O(\frac{1}{n^3})\right) = 1 \]

### Step 3: Expand and simplify

Expanding the left-hand side:

\[ (n + x_n) \left(\frac{1}{n}\right) - (n + x_n) \left(\frac{1}{2n^2}\right) + O\left(\frac{1}{n^3}\right) = 1 \]

This leads to:

\[ 1 + \frac{x_n}{n} - \frac{1}{2n} - \frac{x_n}{2n^2} + O\left(\frac{1}{n^3}\right) = 1 \]

### Step 4: Collect terms

From this, we can collect terms:

\[ \frac{x_n}{n} - \frac{1}{2n} - \frac{x_n}{2n^2} + O\left(\frac{1}{n^3}\right) = 0 \]

### Step 5: Solve for \(x_n\)

As \(n \to \infty\), the dominant term is:

\[ \frac{x_n}{n} - \frac{1}{2n} = 0 \]

Thus,

\[ x_n \approx \frac{1}{2} \quad \text{as } n \to \infty \]

### Step 6: Find the limit

Now we can find the limit:

\[ \lim_{n \to \infty} (2x_n) = 2 \cdot \frac{1}{2} = 1 \]

### Final Answer

Thus, the limit is:

\[ \lim_{n \to \infty} (2x_n) = 1 \]

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