Where is my error in finding the limit $\lim_{n \to \infty} \cos \frac x 2 \cos \frac x 4 \cdots \cos \frac{x}{2^n}$?

I was trying to solve

$$\lim_{n\to\infty} \cos{x\over2}\cos{x\over4}\cos{x\over8}\cdots\cos{x\over2^n}$$

First we can use sine half angle formula from behind and get telescoping series and I got limit $$\lim_{n\to\infty} \cos{x\over2}\cos{x\over4}\cos{x\over8}\cdots\cos{x\over2^n}=\lim_{n\to\infty} \frac{\sin x}{2^n \sin{x\over2^n}}={\sin x\over x}$$

Then I thought using complex number given limit is $$\Re\left[\exp\left(ix\left({1\over2}+{1\over4}+\cdots\right)\right)\right] = \Re[e^{ix}]=\cos x$$

Which answer and solution is correct? What did I do wrong?

Your result with the series assumes that

$$\Re(z_1 z_2) = \Re(z_1) \Re(z_2)$$

for $z_1,z_2 \in \Bbb C$. This is not true.

Let $z_1 = a + bi, z_2 = c + di$. Then $z_1z_2 = (ac - bd) + (ad + bc) i$. The lack of multiplicativity is obvious.

$$ \begin{aligned} \lim _{n \rightarrow \infty} \frac{\sin x}{2^{n} \sin \frac{x}{2^{n}}} &=\lim _{n \rightarrow \infty} \frac{\sin x}{x} \cdot \frac{\frac{x}{2^{n}}}{\sin \left(\frac{x}{2^{n}}\right)} \\ &=\frac{\sin x}{x} \lim _{\frac{x}{2^{n}} \rightarrow 0}\left(\frac{\frac{x}{2^{n}}}{\sin \frac{x}{2^{n}}}\right) \\ &=\frac{\sin x}{x} \end{aligned} $$