`lim_(n rarr oo)((n)/(n^(2)+1^(2))+(n)/(n^(2)+2^(2)) + (n)/(n^(2)+3^(2))+......+(1)/(5n))` is equal to :

Options

A. `(pi)/(2)`
B. `tan^(-1)(3)`
C. `tan^(-1)(2)`
D. `(pi)/(4)`

Explanation

To solve the limit problem step by step, we start with the expression given:

\[ \lim_{n \to \infty} \left( \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \frac{n}{n^2 + 3^2} + \ldots + \frac{n}{5n} \right) \]

### Step 1: Rewrite the last term
The last term can be rewritten as follows:

\[ \frac{1}{5n} = \frac{n}{5n^2} \]

This allows us to express the last term in a form that matches the others.

### Step 2: Generalize the expression
Now, we can express the entire sum as:

\[ \sum_{r=1}^{5} \frac{n}{n^2 + r^2} \]

### Step 3: Factor out \(n^2\)
We can factor \(n^2\) out of the denominator:

\[ \frac{n}{n^2 + r^2} = \frac{n}{n^2(1 + \frac{r^2}{n^2})} = \frac{1}{n(1 + \frac{r^2}{n^2})} \]

### Step 4: Rewrite the limit
Now, substituting this back into the limit, we have:

\[ \lim_{n \to \infty} \sum_{r=1}^{5} \frac{1}{n(1 + \frac{r^2}{n^2})} \]

### Step 5: Recognize the sum as a Riemann sum
As \(n\) approaches infinity, \(\frac{r}{n}\) approaches \(x\) where \(x\) ranges from \(0\) to \(5\). The expression can be interpreted as a Riemann sum for the integral of \(\frac{1}{1+x^2}\) from \(0\) to \(5\).

### Step 6: Convert to integral
Thus, we can write:

\[ \lim_{n \to \infty} \sum_{r=1}^{5} \frac{1}{n(1 + \left(\frac{r}{n}\right)^2)} \approx \int_{0}^{5} \frac{1}{1+x^2} \, dx \]

### Step 7: Calculate the integral
The integral of \(\frac{1}{1+x^2}\) is \(\tan^{-1}(x)\):

\[ \int_{0}^{5} \frac{1}{1+x^2} \, dx = \tan^{-1}(5) - \tan^{-1}(0) = \tan^{-1}(5) - 0 = \tan^{-1}(5) \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \left( \frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \frac{n}{n^2 + 3^2} + \ldots + \frac{n}{5n} \right) = \tan^{-1}(5) \]