We have
`f (x) lim_(n to oo) cos (x)/(2) cos (x)/(2^(2)) cos (x)/(2^(3)) cos (x)/(2^(4))`……
….`cos (x)/(2^(n)) = ("sin" x)/(2^(n) "sin" (x)/(2^(n)))`
using the identity
`lim_(n to oo) lim_(x to 0)` f(x) equals

Options

A. 0
B. 1
C. 2
D. `(1)/(2)`

Answer

To solve the problem step by step, we start with the given expression:

### Step 1: Write the function \( f(x) \)
The function is given as:
\[ f(x) = \lim_{n \to \infty} \left( \cos\left(\frac{x}{2}\right) \cdot \cos\left(\frac{x}{2^2}\right) \cdot \cos\left(\frac{x}{2^3}\right) \cdots \cos\left(\frac{x}{2^n}\right) \right) \]

### Step 2: Recognize the series
We can express this product as a series:
\[ f(x) = \lim_{n \to \infty} \prod_{k=1}^{n} \cos\left(\frac{x}{2^k}\right) \]
This is equivalent to:
\[ f(x) = \lim_{n \to \infty} \sum_{k=1}^{n} \cos\left(\frac{x}{2^k}\right) \]

### Step 3: Use the identity
We are given that:
\[ f(x) = \frac{\sin\left(\frac{x}{2^n}\right)}{2^n \sin\left(\frac{x}{2^n}\right)} \]
This means we can substitute this into our limit:
\[ f(x) = \lim_{n \to \infty} \left( \frac{\sin\left(\frac{x}{2^n}\right)}{2^n \sin\left(\frac{x}{2^n}\right)} \right) \]

### Step 4: Apply the limit
Now we need to evaluate:
\[ \lim_{n \to \infty} \lim_{x \to 0} f(x) \]
Substituting \( f(x) \):
\[ \lim_{n \to \infty} \lim_{x \to 0} \left( \frac{\sin\left(\frac{x}{2^n}\right)}{2^n \sin\left(\frac{x}{2^n}\right)} \right) \]

### Step 5: Use the small angle approximation
Using the limit identity \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2^n}\right)}{\frac{x}{2^n}} = 1 \]
Thus, we can rewrite:
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2^n}\right)}{\sin\left(\frac{x}{2^n}\right)} = 1 \]

### Step 6: Evaluate the limits
Now we can evaluate the limits:
\[ \lim_{n \to \infty} \left( \frac{1}{2^n} \cdot 1 \right) = 0 \]
This means:
\[ \lim_{n \to \infty} \lim_{x \to 0} f(x) = 1 \]

### Final Answer
Thus, the final answer is:
\[ \lim_{n \to \infty} \lim_{x \to 0} f(x) = 1 \]