Question

If $f_{1}(x) = \frac{2}{2+x}$ and $f_{n}(x) = \frac{1}{1+f_{n-1}(x)}$, where $n>1$, then find the approximate value of $f_{50}(1)$.

• A. 0.128
• B. 0.618
• C. 0.666
• D. 0.45

Hint:

When a recursive sequence converges, set $f_n = f_{n-1} = L$ and solve the fixed-point equation. This often avoids long computations.

Answer 1

The Correct Option is B

We are given: \[ f_1(x) = \frac{2}{2+x} \] and for $n>1$, \[ f_n(x) = \frac{1}{1+f_{n-1}(x)} \] Step 1: Compute $f_1(1)$ \[ f_1(1) = \frac{2}{2+1} = \frac{2}{3} \approx 0.6667 \] Step 2: Recursive application The recurrence $f_n(1) = \frac{1}{1+f_{n-1}(1)}$ will converge as $n$ becomes large. Let the limit be $L$. Then: \[ L = \frac{1}{1+L} \] Multiplying both sides by $(1+L)$: \[ L(1+L) = 1 \quad \Rightarrow \quad L^2 + L - 1 = 0 \] Solving: \[ L = \frac{-1 + \sqrt{1+4}}{2} = \frac{-1 + \sqrt{5}}{2} \approx 0.6180 \] Step 3: Conclusion For large $n$, $f_n(1) \to 0.618$. Since $n=50$ is large, $f_{50}(1) \approx 0.618$. \[ \boxed{0.618} \]