The value of `lim_(nrarrinfty) ("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 problem, we need to evaluate the expression:

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

Let’s denote this limit as \( L \):

\[ L = \lim_{n \to \infty} \left( \prod_{r=1}^{n-1} \sin\left(\frac{r\pi}{2n}\right) \right)^{\frac{1}{n}} \]

### Step 1: Take the natural logarithm of \( L \)

Taking the natural logarithm on both sides gives:

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

### Step 2: Convert the sum into an integral

As \( n \to \infty \), the term \( \frac{r}{n} \) can be treated as a continuous variable. We can replace the sum with an integral by letting \( x = \frac{r}{n} \), which implies \( r = nx \) and \( \frac{1}{n} \) becomes \( dx \):

\[ \ln L = \lim_{n \to \infty} \sum_{r=1}^{n-1} \frac{1}{n} \ln\left(\sin\left(\frac{r\pi}{2n}\right)\right) \approx \int_0^1 \ln\left(\sin\left(\frac{\pi x}{2}\right)\right) \, dx \]

### Step 3: Evaluate the integral

The integral we need to evaluate is:

\[ \int_0^1 \ln\left(\sin\left(\frac{\pi x}{2}\right)\right) \, dx \]

To evaluate this integral, we can use a substitution. Let \( t = \frac{\pi x}{2} \), then \( dx = \frac{2}{\pi} dt \) and the limits change from \( x=0 \) to \( x=1 \) to \( t=0 \) to \( t=\frac{\pi}{2} \):

\[ \int_0^1 \ln\left(\sin\left(\frac{\pi x}{2}\right)\right) \, dx = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt \]

### Step 4: Use known results

It is known that:

\[ \int_0^{\frac{\pi}{2}} \ln(\sin t) \, dt = -\frac{\pi}{2} \ln(2) \]

Thus, substituting this result back, we have:

\[ \ln L = \frac{2}{\pi} \left(-\frac{\pi}{2} \ln(2)\right) = -\ln(2) \]

### Step 5: Exponentiate to find \( L \)

Exponentiating both sides gives:

\[ L = e^{-\ln(2)} = \frac{1}{2} \]

### Conclusion

Thus, the value of the limit is:

\[ \boxed{\frac{1}{2}} \]