The value of `lim_(n to oo)sum_(r=1)^(n)(1)/(n) sqrt(((n+r)/(n-r)))` is :

Options

A. `(pi)/(2)-2`
B. `(pi)/(2)+2`
C. `(pi)/(2)-1`
D. `(pi)/(2)+1`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{n+r}{n-r}}, \]

we can follow these steps:

### Step 1: Rewrite the expression inside the limit

We start by rewriting the expression inside the limit:

\[ \sqrt{\frac{n+r}{n-r}} = \sqrt{\frac{n(1 + \frac{r}{n})}{n(1 - \frac{r}{n})}} = \sqrt{\frac{1 + \frac{r}{n}}{1 - \frac{r}{n}}}. \]

Thus, we can rewrite the limit as:

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

### Step 2: Recognize the Riemann sum

As \( n \to \infty \), the term \( \frac{r}{n} \) can be treated as a variable \( x \) where \( x = \frac{r}{n} \). The sum can be interpreted as a Riemann sum for the function \( f(x) = \sqrt{\frac{1+x}{1-x}} \) over the interval from \( 0 \) to \( 1 \):

\[ \sum_{r=1}^{n} \frac{1}{n} f\left(\frac{r}{n}\right) \to \int_0^1 f(x) \, dx. \]

### Step 3: Set up the integral

Thus, we can express our limit as:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \sqrt{\frac{1 + \frac{r}{n}}{1 - \frac{r}{n}}} = \int_0^1 \sqrt{\frac{1+x}{1-x}} \, dx. \]

### Step 4: Solve the integral

To solve the integral

\[ \int_0^1 \sqrt{\frac{1+x}{1-x}} \, dx, \]

we can rationalize the integrand:

\[ \sqrt{\frac{1+x}{1-x}} = \frac{\sqrt{(1+x)(1+x)}}{\sqrt{(1-x)(1+x)}} = \frac{1+x}{\sqrt{1-x^2}}. \]

Thus, the integral becomes:

\[ \int_0^1 \frac{1+x}{\sqrt{1-x^2}} \, dx = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx + \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx. \]

### Step 5: Evaluate the two integrals

1. The first integral:

\[ \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx = \frac{\pi}{2}. \]

2. The second integral can be solved using the substitution \( x = \sin(\theta) \):

\[ \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = \int_0^{\frac{\pi}{2}} \sin(\theta) \, d\theta = 1. \]

### Step 6: Combine the results

Combining both results, we have:

\[ \int_0^1 \sqrt{\frac{1+x}{1-x}} \, dx = \frac{\pi}{2} + 1. \]

### Conclusion

Thus, the final result is:

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

### Final Answer

The value of the limit is:

\[ \frac{\pi}{2} + 1. \]