Let `f(x)=(1)/(1+x)` and let `g(x,n)=f(f(f(….(x))))`, then `lim_(nrarroo)g(x, n)` at x = 1 is

Options

A. `(sqrt5-1)/(2)`
B. `(sqrt5+1)/(2)`
C. `(sqrt3+2)/(2)`
D. `(2-sqrt3)/(2)`

Answer

To solve the problem step by step, we need to analyze the function \( f(x) = \frac{1}{1+x} \) and the nested function \( g(x, n) = f(f(f(...(x)))) \) where \( f \) is applied \( n \) times. We want to find the limit of \( g(x, n) \) as \( n \) approaches infinity, specifically at \( x = 1 \).

### Step 1: Define the function and find its first few iterations
We start with the function:
\[ f(x) = \frac{1}{1+x} \]

Now we compute \( g(x, 1) \):
\[ g(x, 1) = f(x) = \frac{1}{1+x} \]

Next, we compute \( g(x, 2) \):
\[ g(x, 2) = f(f(x)) = f\left(\frac{1}{1+x}\right) = \frac{1}{1 + \frac{1}{1+x}} = \frac{1+x}{2+x} \]

Now compute \( g(x, 3) \):
\[ g(x, 3) = f(g(x, 2)) = f\left(\frac{1+x}{2+x}\right) = \frac{1}{1 + \frac{1+x}{2+x}} = \frac{2+x}{3+2x} \]

### Step 2: Find the limit as \( n \to \infty \)
We want to find:
\[ \lim_{n \to \infty} g(1, n) \]

First, we substitute \( x = 1 \) into our expressions:
- For \( g(1, 1) \):
\[ g(1, 1) = f(1) = \frac{1}{1+1} = \frac{1}{2} \]

- For \( g(1, 2) \):
\[ g(1, 2) = f(f(1)) = f\left(\frac{1}{2}\right) = \frac{1}{1+\frac{1}{2}} = \frac{2}{3} \]

- For \( g(1, 3) \):
\[ g(1, 3) = f(g(1, 2)) = f\left(\frac{2}{3}\right) = \frac{1}{1+\frac{2}{3}} = \frac{3}{5} \]

### Step 3: Establish a pattern and find the limit
We observe that:
- \( g(1, 1) = \frac{1}{2} \)
- \( g(1, 2) = \frac{2}{3} \)
- \( g(1, 3) = \frac{3}{5} \)

It appears that \( g(1, n) \) is approaching a limit. Let's denote this limit as \( y \):
\[ y = \lim_{n \to \infty} g(1, n) \]

From the functional form, we can derive:
\[ y = f(y) = \frac{1}{1+y} \]

Multiplying both sides by \( 1+y \):
\[ y(1+y) = 1 \implies y^2 + y - 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]

Since \( y \) must be positive, we take:
\[ y = \frac{-1 + \sqrt{5}}{2} \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to \infty} g(1, n) = \frac{-1 + \sqrt{5}}{2} \]