The limit `L=lim_(nrarroo)Sigma_(r=1)^(n)(n)/(n^(2)+r^(2))` satisfies

Options

A. `(pi)/(L) lt 6`
B. `(pi)/(L) lt 2`
C. `(pi)/(L) gt 3`
D. `(pi)/(L)lt1`

Answer

To solve the limit \( L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2 + r^2} \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the limit:
\[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2 + r^2} \]
We can factor out \( n^2 \) from the denominator:
\[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2(1 + \frac{r^2}{n^2})} = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n(1 + \frac{r^2}{n^2})} \]

### Step 2: Change of Variables
Let \( x = \frac{r}{n} \). As \( r \) goes from 1 to \( n \), \( x \) goes from \( \frac{1}{n} \) to 1. The sum can be approximated by a Riemann sum:
\[ L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n(1 + \left(\frac{r}{n}\right)^2)} = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n(1 + x^2)} \]

### Step 3: Convert to Integral
As \( n \to \infty \), the sum approaches the integral:
\[ L = \int_{0}^{1} \frac{1}{1 + x^2} \, dx \]

### Step 4: Evaluate the Integral
The integral \( \int \frac{1}{1 + x^2} \, dx \) is known to be:
\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C \]
Thus, we evaluate:
\[ L = \left[ \tan^{-1}(x) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Conclusion
Therefore, the limit \( L \) is:
\[ L = \frac{\pi}{4} \]