Evaluate the following limit:
`lim_(nto oo)(sum_(r=1)^(n) sqrt(r)sum_(r=1)^(h)1/(sqrt(r)))/(sum_(r=1)^(n)r)`

Explanation

To evaluate the limit

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

we will follow these steps:

### Step 1: Rewrite the limit expression
We start by rewriting the limit expression for clarity:

\[ L = \lim_{n \to \infty} \frac{\left( \sum_{r=1}^{n} \sqrt{r} \right) \left( \sum_{r=1}^{n} \frac{1}{\sqrt{r}} \right)}{\sum_{r=1}^{n} r}. \]

### Step 2: Evaluate the sums
Next, we need to evaluate each of the sums as \( n \to \infty \).

1. **Evaluate \( \sum_{r=1}^{n} \sqrt{r} \)**:
The sum \( \sum_{r=1}^{n} \sqrt{r} \) can be approximated by the integral:

\[ \sum_{r=1}^{n} \sqrt{r} \sim \int_{1}^{n} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{n} = \frac{2}{3} (n^{3/2} - 1) \sim \frac{2}{3} n^{3/2}. \]

2. **Evaluate \( \sum_{r=1}^{n} \frac{1}{\sqrt{r}} \)**:
The sum \( \sum_{r=1}^{n} \frac{1}{\sqrt{r}} \) can also be approximated by the integral:

\[ \sum_{r=1}^{n} \frac{1}{\sqrt{r}} \sim \int_{1}^{n} \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_{1}^{n} = 2(\sqrt{n} - 1) \sim 2\sqrt{n}. \]

3. **Evaluate \( \sum_{r=1}^{n} r \)**:
The sum \( \sum_{r=1}^{n} r \) is given by:

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

### Step 3: Substitute the evaluated sums into the limit
Now we substitute the approximations back into the limit:

\[ L = \lim_{n \to \infty} \frac{\left( \frac{2}{3} n^{3/2} \right) \left( 2\sqrt{n} \right)}{\frac{n^2}{2}}. \]

### Step 4: Simplify the expression
Simplifying the expression gives:

\[ L = \lim_{n \to \infty} \frac{\frac{4}{3} n^{3/2} \cdot \sqrt{n}}{\frac{n^2}{2}} = \lim_{n \to \infty} \frac{\frac{4}{3} n^{2}}{\frac{n^2}{2}} = \lim_{n \to \infty} \frac{4}{3} \cdot 2 = \frac{8}{3}. \]

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

\[ \boxed{\frac{8}{3}}. \]