For `n in N,` let `f_(n) (x) = tan ""(x)/(2) (1+ sec x ) (1+ sec 2x) (1+ sec 4x)……(1+ sec 2 ^(n)x),` the `lim _(xto0) (f _(n)(x))/(2x)` is equal to :

Options

A. `0`
B. `2^(n)`
C. `2^(n-1)`
D. `2^(n+1)`

Answer

To solve the problem, we need to find the limit:

\[ \lim_{x \to 0} \frac{f_n(x)}{2x} \]

where

\[ f_n(x) = \frac{\tan(x)}{2(1 + \sec x)(1 + \sec 2x)(1 + \sec 4x) \ldots (1 + \sec 2^n x)}. \]

### Step 1: Rewrite \(f_n(x)\)

We start by rewriting \(f_n(x)\):

\[ f_n(x) = \frac{\tan x}{2(1 + \sec x)(1 + \sec 2x)(1 + \sec 4x) \ldots (1 + \sec 2^n x)}. \]

### Step 2: Use the small angle approximation

As \(x \to 0\), we know that:

\[ \tan x \approx x \quad \text{and} \quad \sec x \approx 1 + \frac{x^2}{2}. \]

Thus, we can approximate:

\[ 1 + \sec kx \approx 1 + \left(1 + \frac{(kx)^2}{2}\right) = 2 + \frac{k^2 x^2}{2}. \]

### Step 3: Substitute the approximations into \(f_n(x)\)

Now substituting this approximation into \(f_n(x)\):

\[ f_n(x) \approx \frac{x}{2 \cdot 2^n \cdot \prod_{k=0}^{n} \left(1 + \frac{(2^k x)^2}{2}\right)}. \]

### Step 4: Simplify the product

The product can be simplified as follows:

\[ \prod_{k=0}^{n} \left(1 + \frac{(2^k x)^2}{2}\right) \approx \prod_{k=0}^{n} \left(1 + \frac{2^{2k} x^2}{2}\right). \]

Using the fact that for small \(x\), \(1 + u \approx e^u\):

\[ \prod_{k=0}^{n} \left(1 + \frac{2^{2k} x^2}{2}\right) \approx e^{\sum_{k=0}^{n} \frac{2^{2k} x^2}{2}} = e^{\frac{x^2}{2} \sum_{k=0}^{n} 2^{2k}}. \]

### Step 5: Calculate the geometric series

The sum of the geometric series is:

\[ \sum_{k=0}^{n} 2^{2k} = \frac{2^{2(n+1)} - 1}{3}. \]

### Step 6: Substitute back into \(f_n(x)\)

Now substituting back, we have:

\[ f_n(x) \approx \frac{x}{2 \cdot 2^n \cdot e^{\frac{x^2}{2} \cdot \frac{2^{2(n+1)} - 1}{3}}}. \]

### Step 7: Find the limit

Now we need to find:

\[ \lim_{x \to 0} \frac{f_n(x)}{2x} = \lim_{x \to 0} \frac{\frac{x}{2 \cdot 2^n}}{2x} = \frac{1}{4 \cdot 2^n}. \]

### Step 8: Final result

Thus, we find:

\[ \lim_{x \to 0} \frac{f_n(x)}{2x} = 2^{n} - 1. \]

### Conclusion

The final answer is:

\[ \lim_{x \to 0} \frac{f_n(x)}{2x} = 2^n - 1. \]