`T h ev a l u eof(lim)_(nvecoo)[t a npi/(2n)tan(2pi)/(2n)dottan(npi)/(2n)]^(1//n)i s` `e` (b) `e^2` (c) 1 (d) `e^3`

Options

A. `e`
B. `e^(2)`
C. `1`
D. `e^(3)`

Explanation

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

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

Let's denote this limit as \( a \):

\[ a = \lim_{n \to \infty} \left( \tan\left(\frac{\pi}{2n}\right) \tan\left(\frac{2\pi}{2n}\right) \cdots \tan\left(\frac{n\pi}{2n}\right) \right)^{\frac{1}{n}} \]

### Step 1: Taking the logarithm

Taking the natural logarithm of both sides, we have:

\[ \log a = \lim_{n \to \infty} \frac{1}{n} \log \left( \tan\left(\frac{\pi}{2n}\right) \tan\left(\frac{2\pi}{2n}\right) \cdots \tan\left(\frac{n\pi}{2n}\right) \right) \]

Using the property of logarithms, we can express this as:

\[ \log a = \lim_{n \to \infty} \frac{1}{n} \left( \log \tan\left(\frac{\pi}{2n}\right) + \log \tan\left(\frac{2\pi}{2n}\right) + \cdots + \log \tan\left(\frac{n\pi}{2n}\right) \right) \]

### Step 2: Expressing the sum in terms of an integral

As \( n \to \infty \), we can approximate the sum by an integral. The argument of the tangent function can be rewritten as:

\[ \frac{k\pi}{2n} \quad \text{for } k = 1, 2, \ldots, n \]

Thus, we can express the sum as:

\[ \sum_{k=1}^{n} \log \tan\left(\frac{k\pi}{2n}\right) \approx n \int_{0}^{1} \log \tan\left(\frac{\pi x}{2}\right) \, dx \]

### Step 3: Evaluating the limit

Substituting this back into our expression for \( \log a \):

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

### Step 4: Evaluating the integral

Now, we need to evaluate the integral:

\[ I = \int_{0}^{1} \log \tan\left(\frac{\pi x}{2}\right) \, dx \]

Using the property of definite integrals, we can also express:

\[ I = \int_{0}^{1} \log \cot\left(\frac{\pi x}{2}\right) \, dx \]

Adding these two integrals gives:

\[ 2I = \int_{0}^{1} \log \tan\left(\frac{\pi x}{2}\right) \, dx + \int_{0}^{1} \log \cot\left(\frac{\pi x}{2}\right) \, dx = \int_{0}^{1} \log(1) \, dx = 0 \]

Thus, we find that:

\[ I = 0 \]

### Step 5: Finding \( a \)

Since \( \log a = 0 \), we have:

\[ a = e^0 = 1 \]

### Conclusion

The value of the limit is:

\[ \boxed{1} \]