`lim_(n to oo)(1)/(n)(1+sqrt((n)/(n+1))+sqrt((n)/(n+2))+....+sqrt((n)/(4n-3)))` is equal to:

Options

A. 1
B. 2
C. 3
D. None of these

Answer

To solve the limit

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \sqrt{\frac{n}{n+1}} + \sqrt{\frac{n}{n+2}} + \ldots + \sqrt{\frac{n}{4n-3}} \right), \]

we can break it down step by step.

### Step 1: Rewrite the expression

The expression inside the limit can be rewritten as:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \sqrt{\frac{n}{n+1}} + \sqrt{\frac{n}{n+2}} + \ldots + \sqrt{\frac{n}{4n-3}} \right). \]

We can express the sum more compactly:

\[ \sum_{r=0}^{3n-3} \sqrt{\frac{n}{n+r}}. \]

### Step 2: Simplifying the square root term

Now, we simplify the term inside the summation:

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

### Step 3: Rewrite the limit

The limit now becomes:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{3n-3} \sqrt{\frac{1}{1 + \frac{r}{n}}}. \]

### Step 4: Change the summation to an integral

As \( n \to \infty \), we can approximate the sum by an integral. We set \( t = \frac{r}{n} \), which implies \( r = nt \) and \( dr = n dt \). The limits of \( r \) from \( 0 \) to \( 3n-3 \) correspond to \( t \) from \( 0 \) to \( 3 - \frac{3}{n} \), which approaches \( 3 \) as \( n \to \infty \).

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} \int_0^3 \sqrt{\frac{1}{1+t}} dt. \]

### Step 5: Evaluate the integral

The integral can be computed as:

\[ \int_0^3 \frac{1}{\sqrt{1+t}} dt. \]

Using the substitution \( u = 1+t \), we have \( du = dt \) and the limits change from \( t=0 \) to \( t=3 \) which corresponds to \( u=1 \) to \( u=4 \):

\[ \int_1^4 \frac{1}{\sqrt{u}} du = 2\sqrt{u} \bigg|_1^4 = 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2. \]

### Final Result

Thus, the limit evaluates to:

\[ \lim_{n \to \infty} \frac{1}{n} \left( 1 + \sqrt{\frac{n}{n+1}} + \sqrt{\frac{n}{n+2}} + \ldots + \sqrt{\frac{n}{4n-3}} \right) = 2. \]

### Conclusion

Therefore, the answer is:

\[ \boxed{2}. \]