if `f(n,theta)=prod_(r=1)^(n) (1-tan^(2)(theta)/(2^(r)))` then the value of `lim_(thetato0) [lim_(ntoinfty) f(n,theta)]`, (where `[x]` represent the greatest integer function) is

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lim_{\theta \to 0} \left[ \lim_{n \to \infty} f(n, \theta) \right] \]

where

\[ f(n, \theta) = \prod_{r=1}^{n} \left( 1 - \frac{\tan^2(\theta)}{2^r} \right) \]

### Step 1: Rewrite the product

We start by rewriting the product \( f(n, \theta) \):

\[ f(n, \theta) = \prod_{r=1}^{n} \left( 1 - \frac{\tan^2(\theta)}{2^r} \right) \]

### Step 2: Analyze the limit as \( n \to \infty \)

As \( n \to \infty \), we consider the behavior of the terms in the product. For small values of \( \theta \), \( \tan(\theta) \approx \theta \), so we can approximate:

\[ f(n, \theta) \approx \prod_{r=1}^{n} \left( 1 - \frac{\theta^2}{2^r} \right) \]

### Step 3: Use logarithms to simplify the product

Taking the logarithm of \( f(n, \theta) \):

\[ \log f(n, \theta) = \sum_{r=1}^{n} \log \left( 1 - \frac{\theta^2}{2^r} \right) \]

For small \( x \), \( \log(1 - x) \approx -x \), so we can approximate:

\[ \log f(n, \theta) \approx -\sum_{r=1}^{n} \frac{\theta^2}{2^r} \]

### Step 4: Evaluate the sum

The sum \( \sum_{r=1}^{n} \frac{1}{2^r} \) is a geometric series:

\[ \sum_{r=1}^{n} \frac{1}{2^r} = 1 - \frac{1}{2^n} \approx 1 \quad \text{as } n \to \infty \]

Thus, we have:

\[ \log f(n, \theta) \approx -\theta^2 \quad \text{as } n \to \infty \]

### Step 5: Exponentiate to find \( f(n, \theta) \)

Exponentiating gives us:

\[ f(n, \theta) \approx e^{-\theta^2} \quad \text{as } n \to \infty \]

### Step 6: Take the limit as \( \theta \to 0 \)

Now we take the limit as \( \theta \to 0 \):

\[ \lim_{\theta \to 0} f(n, \theta) = \lim_{\theta \to 0} e^{-\theta^2} = e^0 = 1 \]

### Step 7: Apply the greatest integer function

Finally, we apply the greatest integer function:

\[ \left[ \lim_{\theta \to 0} \lim_{n \to \infty} f(n, \theta) \right] = [1] = 1 \]

### Final Answer

Thus, the value of

\[ \lim_{\theta \to 0} \left[ \lim_{n \to \infty} f(n, \theta) \right] = 1 \]