`lim_(nrarroo)(sqrt1+sqrt2+…+sqrt(n-1))/(nsqrtn)=`

Options

A. `1//2`
B. `1//3`
C. ``2//3`
D. 0

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \ldots + \sqrt{n-1}}{n \sqrt{n}}, \]

we can follow these steps:

### Step 1: Rewrite the Summation
We can express the numerator as a summation:

\[ \sqrt{1} + \sqrt{2} + \ldots + \sqrt{n-1} = \sum_{r=1}^{n-1} \sqrt{r}. \]

Thus, our limit becomes:

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

### Step 2: Use the Definition of Riemann Sums
As \( n \to \infty \), the sum can be approximated by a definite integral. We can express the sum as:

\[ \sum_{r=1}^{n-1} \sqrt{r} \approx n \int_0^1 \sqrt{x} \, dx, \]

where we substitute \( r = nx \) and \( \Delta x = \frac{1}{n} \).

### Step 3: Change of Variables
Let \( x = \frac{r}{n} \), then \( r = nx \) and \( dr = n \, dx \). The limits change as follows:
- When \( r = 1 \), \( x = \frac{1}{n} \to 0 \) as \( n \to \infty \).
- When \( r = n-1 \), \( x = \frac{n-1}{n} \to 1 \) as \( n \to \infty \).

Thus, the limit can be rewritten as:

\[ \lim_{n \to \infty} \frac{n \int_0^1 \sqrt{nx} \, dx}{n \sqrt{n}}. \]

### Step 4: Evaluate the Integral
Now, we can evaluate the integral:

\[ \int_0^1 \sqrt{nx} \, dx = \sqrt{n} \int_0^1 \sqrt{x} \, dx. \]

The integral \( \int_0^1 \sqrt{x} \, dx \) can be computed as:

\[ \int_0^1 x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_0^1 = \frac{2}{3}. \]

Thus,

\[ \int_0^1 \sqrt{nx} \, dx = \sqrt{n} \cdot \frac{2}{3}. \]

### Step 5: Substitute Back into the Limit
Substituting back, we have:

\[ \lim_{n \to \infty} \frac{n \cdot \sqrt{n} \cdot \frac{2}{3}}{n \sqrt{n}} = \lim_{n \to \infty} \frac{2}{3} = \frac{2}{3}. \]

### Final Answer
Thus, the limit is:

\[ \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \ldots + \sqrt{n-1}}{n \sqrt{n}} = \frac{2}{3}. \]