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 start with the given functions and proceed to find the limit as \( n \) approaches infinity for \( g(x, n) \) at \( x = 1 \).

### Step 1: Define the function \( f(x) \)
We are given:
\[ f(x) = \frac{1}{1+x} \]

### Step 2: Define the function \( g(x, n) \)
The function \( g(x, n) \) is defined as:
\[ g(x, n) = f(f(f(\ldots f(x) \ldots))) \]
where \( f \) is applied \( n \) times.

### Step 3: Evaluate \( g(1, n) \)
We need to find:
\[ g(1, n) = f(f(f(\ldots f(1) \ldots))) \]
Let's start by calculating \( f(1) \):
\[ f(1) = \frac{1}{1+1} = \frac{1}{2} \]

### Step 4: Calculate \( f(f(1)) \)
Next, we find \( f(f(1)) \):
\[ f(f(1)) = f\left(\frac{1}{2}\right) = \frac{1}{1+\frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]

### Step 5: Calculate \( f(f(f(1))) \)
Now we calculate \( f(f(f(1))) \):
\[ f(f(f(1))) = f\left(\frac{2}{3}\right) = \frac{1}{1+\frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \]

### Step 6: Identify the pattern
Continuing this process, we can see a pattern emerging:
- \( g(1, 1) = \frac{1}{2} \)
- \( g(1, 2) = \frac{2}{3} \)
- \( g(1, 3) = \frac{3}{5} \)

It seems that \( g(1, n) \) is approaching a limit as \( n \) increases.

### Step 7: Generalize \( g(1, n) \)
We can generalize the \( n \)-th term:
\[ g(1, n) = \frac{n}{n+1} \]

### Step 8: Find the limit as \( n \to \infty \)
Now we find the limit:
\[ \lim_{n \to \infty} g(1, n) = \lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} = 1 \]

### Conclusion
Thus, the limit we are looking for is:
\[ \lim_{n \to \infty} g(1, n) = 1 \]