`lim_( n to oo) (sin ""(pi)/(2n) . sin ""(2pi)/(2n). sin ""(3pi)/(2n)......sin""((n -1) pi)/(2n))^(1//n)` is equal to

Options

A. `1/2`
B. `1/3`
C. `1/4`
D. None of these

Answer

To solve the limit

\[ \lim_{n \to \infty} \left( \sin\left(\frac{\pi}{2n}\right) \cdot \sin\left(\frac{2\pi}{2n}\right) \cdot \sin\left(\frac{3\pi}{2n}\right) \cdots \sin\left(\frac{(n-1)\pi}{2n}\right) \right)^{\frac{1}{n}}, \]

let's denote this limit as \( y \).

### Step 1: Take the logarithm of both sides
Assuming

\[ y = \lim_{n \to \infty} \left( \sin\left(\frac{\pi}{2n}\right) \cdot \sin\left(\frac{2\pi}{2n}\right) \cdots \sin\left(\frac{(n-1)\pi}{2n}\right) \right)^{\frac{1}{n}}, \]

we take the natural logarithm:

\[ \log y = \lim_{n \to \infty} \frac{1}{n} \log \left( \sin\left(\frac{\pi}{2n}\right) \cdot \sin\left(\frac{2\pi}{2n}\right) \cdots \sin\left(\frac{(n-1)\pi}{2n}\right) \right). \]

### Step 2: Use properties of logarithms
Using the property of logarithms that states \( \log(a \cdot b) = \log a + \log b \), we can rewrite the limit as:

\[ \log y = \lim_{n \to \infty} \frac{1}{n} \left( \log \sin\left(\frac{\pi}{2n}\right) + \log \sin\left(\frac{2\pi}{2n}\right) + \cdots + \log \sin\left(\frac{(n-1)\pi}{2n}\right) \right). \]

### Step 3: Express as a summation
This can be expressed as a summation:

\[ \log y = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n-1} \log \sin\left(\frac{r\pi}{2n}\right). \]

### Step 4: Convert summation to integral
As \( n \to \infty \), the summation can be approximated by an integral:

\[ \log y = \lim_{n \to \infty} \int_{0}^{1} \log \sin\left(\frac{\pi x}{2}\right) \, dx. \]

### Step 5: Change of variables
Let \( t = \frac{\pi x}{2} \), then \( dx = \frac{2}{\pi} dt \). The limits change from \( x = 0 \) to \( x = 1 \) which corresponds to \( t = 0 \) to \( t = \frac{\pi}{2} \):

\[ \log y = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \log \sin t \, dt. \]

### Step 6: Evaluate the integral
The integral \( \int_{0}^{\frac{\pi}{2}} \log \sin t \, dt \) is a known result and equals \( -\frac{\pi}{2} \log 2 \). Thus,

\[ \log y = \frac{2}{\pi} \left(-\frac{\pi}{2} \log 2\right) = -\log 2. \]

### Step 7: Exponentiate to find \( y \)
Exponentiating both sides gives:

\[ y = e^{-\log 2} = \frac{1}{2}. \]

### Final Result
Thus, the limit is:

\[ \lim_{n \to \infty} \left( \sin\left(\frac{\pi}{2n}\right) \cdot \sin\left(\frac{2\pi}{2n}\right) \cdots \sin\left(\frac{(n-1)\pi}{2n}\right) \right)^{\frac{1}{n}} = \frac{1}{2}. \]