Suppose a function f(x) satisfies the following conditions
`f (x+y) =(f(x) +f(y))/(f +f(x) f(y))' AA x , y in R and f '(0)=1`
Also `-1lt f (x) lt 1, AA x in R`
The value of the limit ` l t_(xto oo) (f(x))^(x)` is:

Options

A. 0
B. 1
C. e
D. `e ^(2)`

Answer

To solve the given problem step by step, we will analyze the conditions provided and derive the function \( f(x) \) accordingly.

### Step 1: Understanding the functional equation
We have the functional equation:
\[ f(x+y) = \frac{f(x) + f(y)}{1 + f(x)f(y)} \]
This resembles the addition formula for the tangent function. We will explore if \( f(x) \) can be expressed in terms of the tangent function.

### Step 2: Finding \( f(0) \)
Let's substitute \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f(0+0) = \frac{f(0) + f(0)}{1 + f(0)f(0)} \]
This simplifies to:
\[ f(0) = \frac{2f(0)}{1 + f(0)^2} \]
Let \( f(0) = a \). Then we have:
\[ a = \frac{2a}{1 + a^2} \]
Multiplying both sides by \( 1 + a^2 \):
\[ a(1 + a^2) = 2a \]
This simplifies to:
\[ a + a^3 = 2a \implies a^3 - a = 0 \implies a(a^2 - 1) = 0 \]
Thus, \( a = 0 \) or \( a = 1 \) or \( a = -1 \). Since \( -1 < f(x) < 1 \) for all \( x \), we conclude:
\[ f(0) = 0 \]

### Step 3: Finding the derivative \( f'(x) \)
We know \( f'(0) = 1 \). We can differentiate the functional equation with respect to \( x \) and evaluate at \( x = 0 \):
\[ f'(x+y) = \frac{f'(x)(1 + f(y)f(x)) - f(x)f'(y)f(x)}{(1 + f(x)f(y))^2} \]
Setting \( y = 0 \):
\[ f'(x) = \frac{f'(0)(1 + f(0)f(x)) - f(x)f'(0)f(0)}{(1 + f(0)f(x))^2} \]
Substituting \( f(0) = 0 \) and \( f'(0) = 1 \):
\[ f'(x) = \frac{1 \cdot 1}{1^2} = 1 \]
Thus, \( f'(x) = 1 - f^2(x) \).

### Step 4: Solving the differential equation
We have:
\[ \frac{df}{dx} = 1 - f^2 \]
Separating variables:
\[ \frac{df}{1 - f^2} = dx \]
Integrating both sides:
\[ \frac{1}{2} \ln \left| \frac{1 + f}{1 - f} \right| = x + C \]
Exponentiating gives:
\[ \frac{1 + f}{1 - f} = e^{2(x + C)} = k e^{2x} \]
where \( k = e^{2C} \). Thus,
\[ 1 + f = k e^{2x}(1 - f) \implies 1 + f = k e^{2x} - k e^{2x}f \]
Rearranging gives:
\[ f(1 + k e^{2x}) = k e^{2x} - 1 \]
Thus,
\[ f = \frac{k e^{2x} - 1}{k e^{2x} + 1} \]

### Step 5: Finding the limit
We need to evaluate:
\[ \lim_{x \to \infty} (f(x))^x \]
As \( x \to \infty \), \( e^{2x} \to \infty \):
\[ f(x) = \frac{k e^{2x} - 1}{k e^{2x} + 1} \to 1 \]
This leads to the form \( 1^\infty \), which is indeterminate.

Using the limit property:
\[ \lim_{x \to \infty} f(x)^x = e^{\lim_{x \to \infty} (f(x) - 1)x} \]
Calculating \( f(x) - 1 \):
\[ f(x) - 1 = \frac{k e^{2x} - 1 - (k e^{2x} + 1)}{k e^{2x} + 1} = \frac{-2}{k e^{2x} + 1} \]
Thus,
\[ \lim_{x \to \infty} (f(x) - 1)x = \lim_{x \to \infty} \frac{-2x}{k e^{2x} + 1} = 0 \]
So,
\[ \lim_{x \to \infty} f(x)^x = e^0 = 1 \]

### Final Answer
The value of the limit is:
\[ \boxed{1} \]