The value of `lim_(n to oo)(1)/(n).sum_(r=1)^(2n)(r)/(sqrt(n^(2)+r^(2)))` is equal to

Options

A. `1+sqrt5`
B. `-1+sqrt5`
C. `-1+sqrt2`
D. `1+sqrt2`

Answer

To solve the limit problem given by

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

we will follow a step-by-step approach:

### Step 1: Rewrite the Sum
We start by rewriting the expression inside the limit:

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

### Step 2: Factor Out \(n^2\) from the Square Root
Next, we factor out \(n^2\) from the square root in the denominator:

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

Substituting this back into our limit gives:

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

### Step 3: Change of Variables
Now, we can change variables by letting \(t = \frac{r}{n}\). Then \(r = nt\) and as \(r\) goes from \(1\) to \(2n\), \(t\) goes from \(\frac{1}{n}\) to \(2\). The differential \(dr\) becomes \(n \, dt\).

Thus, we can rewrite the sum as:

\[ \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{nt}{\sqrt{1 + t^2}} \cdot \frac{1}{n} = \lim_{n \to \infty} \sum_{r=1}^{2n} \frac{t}{\sqrt{1 + t^2}} \cdot \Delta t, \]

where \(\Delta t = \frac{1}{n}\).

### Step 4: Convert the Sum to an Integral
As \(n\) approaches infinity, the sum approaches the integral:

\[ \int_{0}^{2} \frac{t}{\sqrt{1 + t^2}} \, dt. \]

### Step 5: Solve the Integral
Now we need to evaluate the integral:

\[ \int \frac{t}{\sqrt{1 + t^2}} \, dt. \]

To solve this, we can use the substitution \(u = 1 + t^2\), which gives \(du = 2t \, dt\) or \(dt = \frac{du}{2t}\). Thus,

\[ \int \frac{t}{\sqrt{1 + t^2}} \, dt = \frac{1}{2} \int u^{-1/2} \, du = \sqrt{1 + t^2} + C. \]

### Step 6: Evaluate the Definite Integral
Now we evaluate the definite integral from \(0\) to \(2\):

\[ \left[ \sqrt{1 + t^2} \right]_{0}^{2} = \sqrt{1 + 2^2} - \sqrt{1 + 0^2} = \sqrt{5} - 1. \]

### Conclusion
Thus, the value of the limit is:

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