`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2 - r^2}}, \]

we will follow these steps:

### Step 1: Rewrite the summation
We start by rewriting the term inside the summation. Notice that we can factor out \( n^2 \) from the square root in the denominator:

\[ \sqrt{n^2 - r^2} = n \sqrt{1 - \left(\frac{r}{n}\right)^2}. \]

Thus, we can rewrite the summation as:

\[ \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2 - r^2}} = \sum_{r=0}^{n-1} \frac{1}{n \sqrt{1 - \left(\frac{r}{n}\right)^2}}. \]

### Step 2: Factor out \( \frac{1}{n} \)
Now, we can factor \( \frac{1}{n} \) out of the summation:

\[ = \frac{1}{n} \sum_{r=0}^{n-1} \frac{1}{\sqrt{1 - \left(\frac{r}{n}\right)^2}}. \]

### Step 3: Convert the summation to an integral
As \( n \to \infty \), the term \( \frac{r}{n} \) approaches \( x \), where \( x \) ranges from 0 to 1 as \( r \) goes from 0 to \( n-1 \). Thus, we can interpret the summation as a Riemann sum for the integral:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \frac{1}{\sqrt{1 - \left(\frac{r}{n}\right)^2}} \approx \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx. \]

### Step 4: Evaluate the integral
The integral

\[ \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx \]

is a standard integral that evaluates to \( \frac{\pi}{2} \). This is because the integral represents the area under the curve of the function \( \frac{1}{\sqrt{1 - x^2}} \), which is the upper half of the unit circle.

### Step 5: Conclusion
Thus, we conclude that:

\[ \lim_{n \to \infty} \sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2 - r^2}} = \frac{\pi}{2}. \]

### Final Answer:
\[ \frac{\pi}{2}. \]