I already solved this problem but I look for more creative and different approaches. Here is my own solution:
Let's define $f_n$ as composition of the function $f$ with itself $n$ times: $f \circ f \circ ... \circ f.$ Here our $f(x) = \frac{x}{2} + 1$ and we want to prove that the limit $f_n(x)$ converges to $2$ as $n$ goes to infinity. My solution was using continued fractions to represent infinite composition.
If we say $y = \frac{\frac{...}{2} + 1}{2} + 1$ and multiply each side with $2$ then subtract $2$ from both sides, we will have $2y - 2 = \frac{...}{2} + 1$ which happens to be $y$ again.
$$2y - 2 = y$$ $$y = 2$$ QED.
So what would be a good alternative to this?
If you compute a few terms of the sequence you can guess what $f_n(x)$ looks like.
Prove by induction that $f_n(x)=\frac x {2^{n}}+\frac 1{2^{n-1}}+\cdots +\frac 1 2 +1$. The limit is $1+\frac 1 2+\frac 1 {2^{2}}+\cdots =2$.