Evaluate
`lim_(nrarroo)(2)/(n)(sin.(pi)/(2n)+sin.(2pi)/(2n)+sin.(3pi)/(2n)+....+sin.(npi)/(2n))`

Answer

To evaluate the limit

\[ \lim_{n \to \infty} \frac{2}{n} \left( \sin\left(\frac{\pi}{2n}\right) + \sin\left(\frac{2\pi}{2n}\right) + \sin\left(\frac{3\pi}{2n}\right) + \ldots + \sin\left(\frac{n\pi}{2n}\right) \right), \]

we can follow these steps:

### Step 1: Rewrite the sum
The sum inside the limit can be expressed as:

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

### Step 2: Use the limit definition of Riemann sums
As \( n \to \infty \), the expression \(\frac{k\pi}{2n}\) approaches 0 for each \( k \). We can relate this sum to a Riemann integral. The term \(\frac{2}{n}\) can be thought of as the width of each subinterval in the Riemann sum, and the sum itself can be approximated by the integral of \(\sin(x)\) from 0 to \(\frac{\pi}{2}\).

### Step 3: Change the variable
Let \( x = \frac{k\pi}{2n} \). Then, \( k = \frac{2nx}{\pi} \) and \( \Delta x = \frac{\pi}{2n} \). The limit can be rewritten as:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \sin\left(\frac{k\pi}{2n}\right) \cdot \frac{2}{n} = \lim_{n \to \infty} \sum_{k=1}^{n} \sin(x) \cdot \Delta x. \]

### Step 4: Evaluate the integral
As \( n \to \infty \), this sum approaches the integral:

\[ \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx. \]

### Step 5: Calculate the integral
The integral of \(\sin(x)\) is:

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

### Step 6: Conclusion
Thus, the limit evaluates to:

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

### Final Answer
The final result is:

\[ \boxed{1}. \]