`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 can 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:

\[ \sqrt{n^2 - r^2} = n \sqrt{1 - \frac{r^2}{n^2}}. \]

Thus, we can rewrite the summation as:

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

### Step 2: Convert the Summation to an Integral
As \(n\) approaches infinity, the term \(\frac{r}{n}\) can be treated as a continuous variable \(x\) in the interval \([0, 1]\). We can express the sum as a Riemann sum:

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

### Step 3: Evaluate the Integral
The integral

\[ \int_{0}^{1} \frac{1}{\sqrt{1 - x^2}} \, dx \]

is a standard integral that evaluates to:

\[ \sin^{-1}(x) \bigg|_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Step 4: 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
The final answer is:

\[ \frac{\pi}{2}. \]

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