Compute `underset(n rarr oo)("lim")(pi)/(n)| "sin"(pi)/(n) + "sin"(2pi)/(n)+ ….+ "sin"((n-1)pi)/(n)|`

Answer

To compute the limit

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

we can follow these steps:

### Step 1: Rewrite the sum
We can express the sum in a more manageable form. Notice that the sum can be rewritten as:

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

Thus, we have:

\[ \lim_{n \to \infty} \frac{\pi}{n} \sum_{k=1}^{n-1} \sin \frac{k\pi}{n}. \]

### Step 2: Recognize the Riemann sum
As \( n \to \infty \), the term \(\frac{1}{n}\) can be interpreted as the width of a small interval in a Riemann sum. We can let \( h = \frac{1}{n} \), and as \( n \to \infty \), \( h \to 0 \). The expression becomes:

\[ \lim_{h \to 0} \sum_{k=1}^{n-1} \sin(k\pi h) \cdot h, \]

where \( k \) goes from \( 1 \) to \( n-1 \).

### Step 3: Convert the sum to an integral
The sum can be approximated by the integral:

\[ \int_0^{1} \sin(\pi x) \, dx, \]

where we have substituted \( x = kh \) and \( k = 1, 2, \ldots, n-1 \).

### Step 4: Evaluate the integral
Now we compute the integral:

\[ \int_0^{1} \sin(\pi x) \, dx. \]

Using the antiderivative of \(\sin(\pi x)\):

\[ \int \sin(\pi x) \, dx = -\frac{1}{\pi} \cos(\pi x) + C. \]

Evaluating the definite integral from \( 0 \) to \( 1 \):

\[ \left[-\frac{1}{\pi} \cos(\pi x)\right]_0^1 = -\frac{1}{\pi} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{\pi} \left( -1 - 1 \right) = \frac{2}{\pi}. \]

### Step 5: Multiply by \(\pi\)
Now, we return to our limit:

\[ \lim_{n \to \infty} \frac{\pi}{n} \sum_{k=1}^{n-1} \sin \frac{k\pi}{n} = \pi \cdot \frac{2}{\pi} = 2. \]

Thus, the final answer is:

\[ \boxed{2}. \]