$lim_{n\rightarrow\infty}\frac{1}{n}\sum_{r=1}^{2n}\frac{r}{\sqrt{n^{2}+r^{2}}}=$

Question

Accepted Answer

To solve the problem and justify why Option 3 is the correct answer, we need to evaluate the limit:

$$
\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{r=1}^{2n}\frac{r}{\sqrt{n^{2}+r^{2}}}
$$

### Step-by-Step Explanation:

1. **Understanding the Expression:**
   The expression inside the sum is $\frac{r}{\sqrt{n^2 + r^2}}$. For large $n$, this can be approximated by considering the dominant terms in the numerator and the denominator.

2. **Approximation for Large $n$:**
   For large $n$, $r$ ranges from 1 to $2n$. We can approximate:
   $$
   \sqrt{n^2 + r^2} \approx n \sqrt{1 + \left(\frac{r}{n}\right)^2}
   $$
   Thus, the term becomes:
   $$
   \frac{r}{\sqrt{n^2 + r^2}} \approx \frac{r}{n \sqrt{1 + \left(\frac{r}{n}\right)^2}} = \frac{\frac{r}{n}}{\sqrt{1 + \left(\frac{r}{n}\right)^2}}
   $$

3. **Converting the Sum to an Integral:**
   As $n \to \infty$, the sum can be approximated by an integral:
   $$
   \frac{1}{n}\sum_{r=1}^{2n}\frac{r}{\sqrt{n^{2}+r^{2}}} \approx \int_{0}^{2} \frac{x}{\sqrt{1 + x^2}} \, dx
   $$
   where $x = \frac{r}{n}$.

4. **Evaluating the Integral:**
   The integral $\int \frac{x}{\sqrt{1 + x^2}} \, dx$ can be solved using a substitution method. Let $u = 1 + x^2$, then $du = 2x \, dx$, and the integral becomes:
   $$
   \int \frac{x}{\sqrt{1 + x^2}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{u}} \, du = \sqrt{u} + C = \sqrt{1 + x^2} + C
   $$
   Evaluating from 0 to 2:
   $$
   \left[\sqrt{1 + x^2}\right]_{0}^{2} = \sqrt{5} - 1
   $$

5. **Conclusion:**
   The result of the integral is $\sqrt{5} - 1$, which matches Option 3: $-1 + \sqrt{5}$.

### Explanation of Incorrect Options:

- **Option 1: $-1 + \sqrt{2}$**
  - This option suggests a different integral evaluation, which does not match the correct calculation of $\sqrt{5} - 1$.

- **Option 2: $1 + \sqrt{2}$**
  - This option is incorrect because the integral evaluation does not yield a result involving $\sqrt{2}$.

- **Option 4: $1 + \sqrt{5}$**
  - This option is incorrect because the integral evaluation results in $\sqrt{5} - 1$, not $1 + \sqrt{5}$.

Thus, the correct answer is indeed Option 3: $-1 + \sqrt{5}$.