`lim_(n->oo) co sx/2*co sx/4*co sx/8......co sx/(2^n)`

Answer

To solve the limit \( \lim_{n \to \infty} \cos\left(\frac{x}{2}\right) \cos\left(\frac{x}{4}\right) \cos\left(\frac{x}{8}\right) \ldots \cos\left(\frac{x}{2^n}\right) \), we can follow these steps:

### Step 1: Rewrite the product using sine
We know from trigonometric identities that:
\[ \sin(2x) = 2 \sin(x) \cos(x) \]
This can be rearranged to express sine in terms of cosine:
\[ \sin(x) = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]
We can apply this identity iteratively.

### Step 2: Apply the identity repeatedly
Applying the identity repeatedly, we can express the sine function in terms of cosines:
\[ \sin(x) = 2^n \sin\left(\frac{x}{2^n}\right) \prod_{k=0}^{n-1} \cos\left(\frac{x}{2^k}\right) \]
This means:
\[ \prod_{k=0}^{n-1} \cos\left(\frac{x}{2^k}\right) = \frac{\sin(x)}{2^n \sin\left(\frac{x}{2^n}\right)} \]

### Step 3: Take the limit as \( n \to \infty \)
Now, we need to evaluate:
\[ \lim_{n \to \infty} \prod_{k=0}^{n-1} \cos\left(\frac{x}{2^k}\right) = \lim_{n \to \infty} \frac{\sin(x)}{2^n \sin\left(\frac{x}{2^n}\right)} \]

### Step 4: Analyze the limit
As \( n \to \infty \), \( \frac{x}{2^n} \to 0 \). We can use the fact that:
\[ \lim_{y \to 0} \frac{\sin(y)}{y} = 1 \]
Thus, \( \sin\left(\frac{x}{2^n}\right) \sim \frac{x}{2^n} \) as \( n \to \infty \). Therefore, we have:
\[ \sin\left(\frac{x}{2^n}\right) \approx \frac{x}{2^n} \]

### Step 5: Substitute back into the limit
Substituting this approximation back, we get:
\[ \lim_{n \to \infty} \frac{\sin(x)}{2^n \cdot \frac{x}{2^n}} = \lim_{n \to \infty} \frac{\sin(x)}{x} = \frac{\sin(x)}{x} \]

### Final Result
Thus, the final result is:
\[ \lim_{n \to \infty} \cos\left(\frac{x}{2}\right) \cos\left(\frac{x}{4}\right) \cos\left(\frac{x}{8}\right) \ldots \cos\left(\frac{x}{2^n}\right) = \frac{\sin(x)}{x} \]