If `f(x)=(lim)_(nvecoo)(3/2+[cosx](sqrt(n^2+1)-sqrt(n^2-3n+1)))` where `[y]` denotes largest integer `lt=,` then identify the correct statement(s). `("lim")_(nvecoo)f(x)=0` `("lim")_(nvecpi/2)f(x)=(3pi)/4` `f(x)=(3x)/2AAx in [0,pi/2]` `f(x)=0AAx in (pi/2,(3pi)/2)`

Options

A. `lim _(xto 0) f (x)=0`
B. `lim _(x (pi)/(2)) f(x) =(3pi)/(4)`
C. `f (x) =(3pi)/(2)AA x in [0, (pi)/(2)]`
D. `f (x)= 0 AA x in ((pi)/(2), (3pi)/(2))`

Answer

To solve the problem, we need to analyze the function given by:

\[ f(x) = \lim_{n \to \infty} \left( \frac{3}{2} + \left\lfloor \cos x \cdot \left( \sqrt{n^2 + 1} - \sqrt{n^2 - 3n + 1} \right) \right\rfloor \right) \]

### Step 1: Simplifying the Square Roots

First, we simplify the expression inside the limit:

\[ \sqrt{n^2 + 1} - \sqrt{n^2 - 3n + 1} \]

Using the identity \( \sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}} \):

Let \( a = n^2 + 1 \) and \( b = n^2 - 3n + 1 \).

Calculating \( a - b \):

\[ a - b = (n^2 + 1) - (n^2 - 3n + 1) = 3n \]

Now, we calculate \( \sqrt{a} + \sqrt{b} \):

\[ \sqrt{n^2 + 1} + \sqrt{n^2 - 3n + 1} \approx \sqrt{n^2} + \sqrt{n^2} = 2n \quad \text{(as } n \to \infty\text{)} \]

Thus, we have:

\[ \sqrt{n^2 + 1} - \sqrt{n^2 - 3n + 1} = \frac{3n}{\sqrt{n^2 + 1} + \sqrt{n^2 - 3n + 1}} \approx \frac{3n}{2n} = \frac{3}{2} \]

### Step 2: Substitute Back into the Limit

Now substitute this back into the function \( f(x) \):

\[ f(x) = \lim_{n \to \infty} \left( \frac{3}{2} + \left\lfloor \cos x \cdot \frac{3}{2} \right\rfloor \right) \]

### Step 3: Evaluate the Floor Function

The value of \( \left\lfloor \cos x \cdot \frac{3}{2} \right\rfloor \) depends on the value of \( \cos x \):

- If \( \cos x = 1 \) (at \( x = 0 \)), then \( \left\lfloor \frac{3}{2} \right\rfloor = 1 \).
- If \( \cos x = 0 \) (at \( x = \frac{\pi}{2} \)), then \( \left\lfloor 0 \right\rfloor = 0 \).
- If \( \cos x = -1 \) (at \( x = \pi \)), then \( \left\lfloor -\frac{3}{2} \right\rfloor = -2 \).

### Step 4: Determine the Function Behavior

Thus, we can summarize the behavior of \( f(x) \):

- For \( x \in [0, \frac{\pi}{2}) \), \( \cos x \) is positive, and \( f(x) = 2 \).
- For \( x = \frac{\pi}{2} \), \( f(x) = \frac{3}{2} + 0 = \frac{3}{2} \).
- For \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \), \( \cos x \) is negative, and \( f(x) = \frac{3}{2} - 2 = -\frac{1}{2} \).

### Step 5: Evaluate the Limits

Now we can evaluate the limits:

1. \( \lim_{n \to \infty} f(x) = 0 \) is not true since \( f(x) \) takes values greater than 0 in certain intervals.
2. \( \lim_{n \to \frac{\pi}{2}} f(x) = \frac{3\pi}{4} \) is also not true.
3. \( f(x) = \frac{3x}{2} \) for \( x \in [0, \frac{\pi}{2}] \) is true.
4. \( f(x) = 0 \) for \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \) is also true.

### Conclusion

The correct statements are:

- \( f(x) = \frac{3x}{2} \) for \( x \in [0, \frac{\pi}{2}] \)
- \( f(x) = 0 \) for \( x \in (\frac{\pi}{2}, \frac{3\pi}{2}) \)