If `x_1=sqrt(3) and x_(n+1)=(x_n)/(1+sqrt(1+x_ n^2)),AA n in N` then `lim_(n->oo)2^n x_n` is equal to

Options

A. `3/(2pi)`
B. `2/(3pi)`
C. `(3pi)/3`
D. `(3pi)/2`

Answer

To solve the problem, we need to find the limit of \( 2^n x_n \) as \( n \) approaches infinity, given the recursive sequence defined by \( x_1 = \sqrt{3} \) and \( x_{n+1} = \frac{x_n}{1 + \sqrt{1 + x_n^2}} \).

Let's go through the steps:

### Step 1: Calculate \( x_2 \)
Using the recursive formula:
\[ x_2 = \frac{x_1}{1 + \sqrt{1 + x_1^2}} \]
Substituting \( x_1 = \sqrt{3} \):
\[ x_2 = \frac{\sqrt{3}}{1 + \sqrt{1 + (\sqrt{3})^2}} = \frac{\sqrt{3}}{1 + \sqrt{1 + 3}} = \frac{\sqrt{3}}{1 + \sqrt{4}} = \frac{\sqrt{3}}{1 + 2} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]

### Step 2: Calculate \( x_3 \)
Now, we calculate \( x_3 \):
\[ x_3 = \frac{x_2}{1 + \sqrt{1 + x_2^2}} = \frac{\frac{1}{\sqrt{3}}}{1 + \sqrt{1 + \left(\frac{1}{\sqrt{3}}\right)^2}} = \frac{\frac{1}{\sqrt{3}}}{1 + \sqrt{1 + \frac{1}{3}}} = \frac{\frac{1}{\sqrt{3}}}{1 + \sqrt{\frac{4}{3}}} = \frac{\frac{1}{\sqrt{3}}}{1 + \frac{2}{\sqrt{3}}} \]
Simplifying further:
\[ = \frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3} + 2}{\sqrt{3}}} = \frac{1}{\sqrt{3}(\sqrt{3} + 2)} = \frac{1}{2 + \sqrt{3}} \]

### Step 3: Identify a pattern
We observe the values:
- \( x_1 = \sqrt{3} \)
- \( x_2 = \frac{1}{\sqrt{3}} \)
- \( x_3 = \frac{1}{2 + \sqrt{3}} \)

It appears that \( x_n \) can be expressed in terms of \( \tan \) functions:
- \( x_1 = \tan\left(\frac{\pi}{3}\right) \)
- \( x_2 = \tan\left(\frac{\pi}{6}\right) \)
- \( x_3 = \tan\left(\frac{\pi}{12}\right) \)

This suggests that:
\[ x_n = \tan\left(\frac{\pi}{3 \cdot 2^{n-1}}\right) \]

### Step 4: Find \( 2^n x_n \)
Now we compute:
\[ 2^n x_n = 2^n \tan\left(\frac{\pi}{3 \cdot 2^{n-1}}\right) \]
As \( n \to \infty \), \( \frac{\pi}{3 \cdot 2^{n-1}} \to 0 \). We can use the small angle approximation \( \tan x \approx x \) when \( x \) is small:
\[ \tan\left(\frac{\pi}{3 \cdot 2^{n-1}}\right) \approx \frac{\pi}{3 \cdot 2^{n-1}} \]
Thus:
\[ 2^n x_n \approx 2^n \cdot \frac{\pi}{3 \cdot 2^{n-1}} = \frac{2 \pi}{3} \]

### Step 5: Take the limit
Finally, we take the limit:
\[ \lim_{n \to \infty} 2^n x_n = \frac{2 \pi}{3} \]

### Conclusion
The limit is:
\[ \lim_{n \to \infty} 2^n x_n = \frac{2\pi}{3} \]