`underset(nrarroo)"lim"[sin'(pi)`/(n)+sin'(2pi)`/(n)+"......"+sin((n-1))/(n)`pi`] is equal to :

Options

A. 0
B. `pi`
C. `oo`
D. none of these

Answer

To solve the limit

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

we can use the formula for the sum of sines. The sum can be expressed as:

\[ \sum_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right). \]

### Step 1: Use the formula for the sum of sines

The formula for the sum of sines is given by:

\[ \sum_{k=1}^{m} \sin(k\theta) = \frac{\sin\left(\frac{m\theta}{2}\right) \sin\left(\frac{(m+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}. \]

In our case, \( m = n-1 \) and \( \theta = \frac{\pi}{n} \). Thus, we can write:

\[ \sum_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = \frac{\sin\left(\frac{(n-1)\pi}{2n}\right) \sin\left(\frac{n\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}. \]

### Step 2: Simplify the expression

Now we can simplify the expression:

\[ = \frac{\sin\left(\frac{(n-1)\pi}{2n}\right) \sin\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2n}\right)}. \]

Since \( \sin\left(\frac{\pi}{2}\right) = 1 \), we have:

\[ = \frac{\sin\left(\frac{(n-1)\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}. \]

### Step 3: Evaluate the limit

Now we need to evaluate the limit as \( n \to \infty \):

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

As \( n \to \infty \), both \( \frac{(n-1)\pi}{2n} \) and \( \frac{\pi}{2n} \) approach \( 0 \). Therefore, we can use the small angle approximation \( \sin(x) \approx x \) when \( x \) is small:

\[ \sin\left(\frac{(n-1)\pi}{2n}\right) \approx \frac{(n-1)\pi}{2n} \quad \text{and} \quad \sin\left(\frac{\pi}{2n}\right) \approx \frac{\pi}{2n}. \]

Substituting these approximations into our limit gives:

\[ \lim_{n \to \infty} \frac{\frac{(n-1)\pi}{2n}}{\frac{\pi}{2n}} = \lim_{n \to \infty} \frac{(n-1)}{1} = \lim_{n \to \infty} (n-1) = \infty. \]

### Final Result

Thus, the limit evaluates to:

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