The value of `lim_(nto oo)(sqrt(n^(2)+n+1)-[sqrt(n^(2)+n+1)])` where [.] denotes the greatest integer function is

Options

A. 0
B. `1//2`
C. `2//3`
D. `1//4`

Answer

To solve the limit \( \lim_{n \to \infty} \left( \sqrt{n^2 + n + 1} - \left[ \sqrt{n^2 + n + 1} \right] \right) \), where \([.]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Analyze the expression inside the limit
We start with the expression \( \sqrt{n^2 + n + 1} \). As \( n \) approaches infinity, we can approximate this expression.

### Step 2: Approximate \( \sqrt{n^2 + n + 1} \)
We can factor out \( n^2 \) from the square root:
\[ \sqrt{n^2 + n + 1} = \sqrt{n^2(1 + \frac{1}{n} + \frac{1}{n^2})} = n\sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} \]

### Step 3: Expand the square root using Taylor series
Using the binomial expansion for large \( n \):
\[ \sqrt{1 + x} \approx 1 + \frac{x}{2} \text{ for small } x \]
We can apply this to our expression:
\[ \sqrt{1 + \frac{1}{n} + \frac{1}{n^2}} \approx 1 + \frac{1/2}{n} + \text{(higher order terms)} \]
Thus,
\[ \sqrt{n^2 + n + 1} \approx n\left(1 + \frac{1/2}{n}\right) = n + \frac{1}{2} \]

### Step 4: Determine the greatest integer function
Now, since \( \sqrt{n^2 + n + 1} \approx n + \frac{1}{2} \), we can conclude:
\[ \left[ \sqrt{n^2 + n + 1} \right] = n \text{ for large } n \]

### Step 5: Substitute back into the limit
Now we substitute back into the limit:
\[ \lim_{n \to \infty} \left( \sqrt{n^2 + n + 1} - n \right) = \lim_{n \to \infty} \left( n + \frac{1}{2} - n \right) = \lim_{n \to \infty} \frac{1}{2} = \frac{1}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{1}{2} \]