The value of `lim_(nrarroo)(cos x cos .(x)/(2)cos.(x)/(4)..cos .(x)/(2^(n)))` is equal to

Options

A. `(x)/(sinx)`
B. `(sinx)/(x)`
C. `(sin2x)/(2x)`
D. `(2x)/(sin2x)`

Answer

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

### Step 1: Rewrite the limit
We start by rewriting the expression in a more manageable form:
\[ \lim_{n \to \infty} \prod_{k=0}^{n} \cos \frac{x}{2^k} \]
This indicates that we are taking the product of cosines from \( k = 0 \) to \( n \).

### Step 2: Use the product-to-sum identity
We can utilize the identity for the product of cosines:
\[ \prod_{k=0}^{n} \cos \frac{x}{2^k} = \frac{\sin \left( \frac{x}{2^{n+1}} \right)}{2^{n+1} \sin \left( \frac{x}{2} \right)} \]
This identity helps us express the product of cosines in terms of sine.

### Step 3: Substitute into the limit
Substituting this identity into our limit gives:
\[ \lim_{n \to \infty} \frac{\sin \left( \frac{x}{2^{n+1}} \right)}{2^{n+1} \sin \left( \frac{x}{2} \right)} \]

### Step 4: Evaluate the limit
As \( n \to \infty \), \( \frac{x}{2^{n+1}} \) approaches 0. Therefore, we can use the small angle approximation \( \sin y \approx y \) when \( y \) is close to 0:
\[ \sin \left( \frac{x}{2^{n+1}} \right) \approx \frac{x}{2^{n+1}} \]
Thus, we have:
\[ \lim_{n \to \infty} \frac{\frac{x}{2^{n+1}}}{2^{n+1} \sin \left( \frac{x}{2} \right)} = \lim_{n \to \infty} \frac{x}{(2^{n+1})^2 \sin \left( \frac{x}{2} \right)} = 0 \]
since \( (2^{n+1})^2 \) grows without bound as \( n \to \infty \).

### Final Result
Therefore, the value of the limit is:
\[ \lim_{n \to \infty} \left( \cos x \cdot \cos \frac{x}{2} \cdots \cos \frac{x}{2^n} \right) = 0 \]