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

Options

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

Answer

To solve the limit

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

we can break it down step by step.

### Step 1: Rewrite the expression

The limit can be rewritten as:

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

### Step 2: Simplify each term

Each term in the summation can be simplified:

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

### Step 3: Factor out \(n\)

Now, we can factor out \(n\) from the limit:

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

### Step 4: Change the summation to an integral

As \(n\) approaches infinity, \(\frac{1}{n}\) can be treated as \(dx\) in the integral form. The term \(\frac{r}{n}\) approaches \(x\) as \(n\) becomes very large. Thus, we can express the summation as an integral:

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

### Step 5: Evaluate the integral

The integral

\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x). \]

Thus, we evaluate:

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

### Final Result

Therefore, the limit evaluates to:

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