`lim_(n to oo ) {(n)/(n^(2)+1^(2))+(n)/(n^(2)+2^(2))+....+ (n)/(n^(2)+n^(2))}` is equal to

Options

A. 1
B. 0
C. `(pi)/(4)`
D. `(pi)/(2)`

Explanation

To solve the limit

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

we can rewrite the expression using summation notation:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2 + r^2}. \]

### Step 1: Rewrite the terms in the summation
We can factor \(n^2\) out of the denominator:

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

### Step 2: Substitute into the summation
Now we can rewrite the limit as:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \cdot \frac{1}{1 + \left(\frac{r}{n}\right)^2}. \]

### Step 3: Recognize the Riemann sum
As \(n \to \infty\), the term \(\frac{r}{n}\) approaches a continuous variable \(x\) ranging from \(0\) to \(1\). Thus, the summation can be interpreted as a Riemann sum for the integral:

\[ \int_{0}^{1} \frac{1}{1 + x^2} \, dx. \]

### Step 4: Evaluate the integral
The integral \(\int \frac{1}{1 + x^2} \, dx\) is known to be:

\[ \tan^{-1}(x). \]

Evaluating this from \(0\) to \(1\):

\[ \left[ \tan^{-1}(x) \right]_{0}^{1} = \tan^{-1}(1) - \tan^{-1}(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}. \]

### Final Result
Thus, we conclude that:

\[ \lim_{n \to \infty} \left( \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \ldots + \frac{n}{n^2 + n^2} \right) = \frac{\pi}{4}. \]