Evaluate `lim_(ntooo)ncos((pi)/(4n))sin((pi)/(4n)).`

Explanation

To evaluate the limit \( \lim_{n \to \infty} n \cos\left(\frac{\pi}{4n}\right) \sin\left(\frac{\pi}{4n}\right) \), we can follow these steps:

### Step 1: Change the variable
Let \( \theta = \frac{\pi}{4n} \). As \( n \to \infty \), \( \theta \to 0 \).

### Step 2: Rewrite the limit
We can express \( n \) in terms of \( \theta \):
\[ n = \frac{\pi}{4\theta} \]
Now, rewrite the limit:
\[ \lim_{n \to \infty} n \cos\left(\frac{\pi}{4n}\right) \sin\left(\frac{\pi}{4n}\right) = \lim_{\theta \to 0} \frac{\pi}{4\theta} \cos(\theta) \sin(\theta) \]

### Step 3: Use the double angle identity
Using the identity \( 2 \cos(\theta) \sin(\theta) = \sin(2\theta) \), we rewrite the expression:
\[ \lim_{\theta \to 0} \frac{\pi}{4\theta} \cdot \frac{1}{2} \sin(2\theta) = \lim_{\theta \to 0} \frac{\pi}{8\theta} \sin(2\theta) \]

### Step 4: Apply the limit
Now, we can apply the limit:
\[ \lim_{\theta \to 0} \frac{\sin(2\theta)}{2\theta} = 1 \]
Thus, we have:
\[ \lim_{\theta \to 0} \frac{\pi}{8\theta} \sin(2\theta) = \frac{\pi}{8} \cdot 1 = \frac{\pi}{8} \]

### Step 5: Final result
Therefore, the final result is:
\[ \lim_{n \to \infty} n \cos\left(\frac{\pi}{4n}\right) \sin\left(\frac{\pi}{4n}\right) = \frac{\pi}{8} \]