For which values of $x$ the limit $\displaystyle\lim_{n\to \infty} (-1)^n \frac x{2^n}$ exists?

Question

Let $f_1:[0,4]\to [0,4] $be defined by $f_1(x)=3-\frac x2$. Define $f_n(x) =f_1\left( f_{n-1}(x)\right)$ for $n\ge 2$. Find the set of all $x$ such that $\displaystyle\lim_{n\to \infty} f_n(x)$ exists and also find the corresponding limits.

Try
I find that $$f_n(x) =\frac 3{2^{n-1}}+(-1)^n \frac x{2^n}$$
Hence $$\displaystyle\lim_{n\to \infty} f_n(x)=\displaystyle\lim_{n\to \infty} \frac 3{2^{n-1}}+\displaystyle\lim_{n\to \infty} (-1)^n \frac x{2^n} =0+\displaystyle\lim_{n\to \infty} (-1)^n \frac x{2^n}=\displaystyle\lim_{n\to \infty} (-1)^n \frac x{2^n}$$ Hence the left hand side exists if the right hand side is so. From here I'm unable to find the values of $x$.
Please help. Thanks in advance

Answer 1

$$f_n(x) = 2 \left[ 1- \left( -\frac12\right)^n\right]+(-1)^n \frac{x}{2^n}$$

For any finite number $x$, $$\lim_{n\to \infty} (-1)^n \frac{x}{2^n}=0$$since $2^n \to \infty$.

Your domain is $[0,4]$, everyone of them would satisfy the condition.

Note the first term of your $f_n$ is not right.

Answer 2

It is easy to show that $$\lim_{n\to\infty} \frac{(-1)^n}{2^n} = 0,$$ and since $x$ is independent of $n$, $\lim_{n\to\infty} x = x,$ so

$$ \lim_{n\to\infty} (-1)^n \frac x{2^n} = \left(\lim_{n\to\infty} \frac{(-1)^n}{2^n}\right) \left(\lim_{n\to\infty} x\right) = 0 \cdot x = 0.$$

On the other hand,

\begin{align} 3 - \frac12\left(\frac 3{2^{n-1}}+(-1)^n \frac x{2^n}\right) &= 3 - \left(\frac 3{2^n}+(-1)^n \frac x{2^{n+1}}\right) \\ &= 3 - \frac 3{2^n} + (-1)^{n+1} \frac x{2^{n+1}} \\ &\neq \frac 3{2^n} + (-1)^{n+1} \frac x{2^{n+1}} \end{align}

for any $n > 1.$ So you have some corrections to make in the calculations that got you your formula.