Which of the following is/are unity? (where [.] and {.} denote the greatest integer and fractional part functions, respectively)

Options

A. `lim_(xrarroo)sin((x^(2)+2)/(x^(2)+1))`
B. `lim_(xrarr0)[(x^(2)+2)/(x^(2)+1)]`
C. `lim_(xrarroo){(x^(2)+2)/(x^(2)+1)}`
D. `lim_(xrarroo)((x^(2))/(x^(2)+1))`

Answer

To determine which of the given options is/are unity, we will analyze each option step by step.

### Step-by-Step Solution:

**Option 1:**
Evaluate the limit as \( x \) approaches infinity for the expression \( \frac{\sin(x^2) + 2}{x^2 + 1} \).

1. **Identify the limit form:**
\[ \lim_{x \to \infty} \frac{\sin(x^2) + 2}{x^2 + 1} \]
As \( x \) approaches infinity, both the numerator and denominator approach infinity.

2. **Apply L'Hôpital's Rule:**
Since we have an indeterminate form \( \frac{\infty}{\infty} \), we can apply L'Hôpital's Rule:
\[ \lim_{x \to \infty} \frac{\sin(x^2) + 2}{x^2 + 1} = \lim_{x \to \infty} \frac{2x \cos(x^2)}{2x} = \lim_{x \to \infty} \cos(x^2) \]
The limit of \( \cos(x^2) \) oscillates between -1 and 1.

3. **Conclusion for Option 1:**
Since \( \cos(x^2) \) does not converge to a single value, this limit does not equal 1. Therefore, **Option 1 is not unity**.

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**Option 2:**
Evaluate the limit as \( x \) approaches 0 for the expression \( \lfloor \frac{x^2 + 2}{x^2 + 1} \rfloor \).

1. **Calculate the limit:**
\[ \lim_{x \to 0} \frac{x^2 + 2}{x^2 + 1} = \frac{0 + 2}{0 + 1} = 2 \]
Therefore, \( \lfloor 2 \rfloor = 2 \).

2. **Conclusion for Option 2:**
Since the limit is 2, **Option 2 is not unity**.

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**Option 3:**
Evaluate the limit as \( x \) approaches 0 for the expression \( \{ x \} \) where \( \{ x \} = x - \lfloor x \rfloor \).

1. **Calculate the limit:**
As \( x \) approaches 0, \( \lfloor x \rfloor = 0 \), thus:
\[ \{ x \} = x - 0 = x \]
Therefore, \( \lim_{x \to 0} \{ x \} = 0 \).

2. **Conclusion for Option 3:**
Since the limit is 0, **Option 3 is not unity**.

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**Option 4:**
Evaluate the limit as \( x \) approaches infinity for the expression \( \frac{x^2}{x^2 + 1} \).

1. **Calculate the limit:**
\[ \lim_{x \to \infty} \frac{x^2}{x^2 + 1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x^2}} = \frac{1}{1 + 0} = 1 \]

2. **Conclusion for Option 4:**
Since the limit is 1, **Option 4 is unity**.

### Final Conclusion:
Among the options analyzed, only **Option 4 is unity**.

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