Limiting Recurrence

From given conditions, we write $\begin{aligned} &f_2 (x) = \frac{x}{4} + \frac{30}{2} \\ &f_3 (x) = \frac{x}{8} + \frac{70}{4} \\ &f_4 (x) = \frac{x}{16} + \frac{150}{8} \\ &\text{and from above we write the general term as:} \\ &f_n (x) = \frac{x}{2^{n}} + \frac{10+20+40+80+160+320......+10\times 2^{n-1}}{2^{n-1}} \\ & = \frac{x}{2^{n}} + \frac{10\times(1+2+4+8+16+32......+ 2^{n-1})}{2^{n-1}}\\ & = \frac{x}{2^{n}} + \frac{10\times(2^{n} - 1 )}{2^{n-1}} \\ & = \frac{x}{2^{n}} + 20 - \frac{10}{2^{n-1}} \\ & \Rightarrow \lim_{n\to\infty} f(x) = 20 \end{aligned}$

i will share an easier approach

As n tends to infinity n tends to n-1

therefore

fn(x) = f1(fn(x))

substitute values

fn(x) = fn(x)/2+10

therefore fn(x) = 20

Relevant wiki: Linear Recurrence Relations - With Repeated Roots

$\begin{aligned} f_1 (x) & = \frac x2 + 10 \\ f_n (x) & = f_1 \left(f_{n-1}(x)\right) \\ \implies f_n (x) & = \frac {f_{n-1}(x)}2 + 10 \\ f_n (x) - 20 & = \frac 12 \left(f_{n-1}(x) - 20 \right) & \small \color{#3D99F6} \text{Let }g_n = f_n(x) - 20 \\ \implies g_n & = \frac 12 g_{n-1} & \small \color{#3D99F6} \text{Characteristic equation: }2r-1=0 \implies r = \frac 12 \\ g_n & = \frac c{2^n} \\ g_1 & = \frac c2 & \small \color{#3D99F6} \text{Note that }g_1 = f_1(x) - 20 = \frac x2 - 10 \\ \frac c2 & = \frac x2 - 10 \\ \implies c & = x - 20 \\ \implies g_n & = \frac {x - 20}{2^n} \\ f_n (x) - 20 & = \frac {x-20}{2^n} \\ \implies f_n(x) & = \frac {x-20}{2^n} + 20 \\ \implies \lim_{n \to \infty} f_n(x) & = \boxed{20} \end{aligned}$