Solve (i) `lim_(xtooo)tan^(-1)x`
(ii) `lim_(xto-oo)tan^(-1)x`
(where [.] denotes greatest integer function)

Answer

To solve the limits given in the question, we will analyze the behavior of the function \( \tan^{-1}(x) \) as \( x \) approaches infinity and negative infinity.

### Step-by-Step Solution:

**(i) Calculate \( \lim_{x \to \infty} \tan^{-1}(x) \)**

1. **Understanding the Function**: The function \( \tan^{-1}(x) \) is the inverse tangent function. It is defined for all real numbers and has horizontal asymptotes.

2. **Behavior as \( x \to \infty \)**: As \( x \) increases towards infinity, the value of \( \tan^{-1}(x) \) approaches \( \frac{\pi}{2} \). This is because the tangent function approaches infinity as its input approaches \( \frac{\pi}{2} \).

3. **Conclusion**: Therefore, we can write:
\[ \lim_{x \to \infty} \tan^{-1}(x) = \frac{\pi}{2} \]

4. **Applying the Greatest Integer Function**: Since we need the greatest integer function of this limit:
\[ \lfloor \lim_{x \to \infty} \tan^{-1}(x) \rfloor = \lfloor \frac{\pi}{2} \rfloor \]
The value of \( \frac{\pi}{2} \) is approximately \( 1.57 \), so:
\[ \lfloor \frac{\pi}{2} \rfloor = 1 \]

**Final Answer for (i)**:
\[ \lfloor \lim_{x \to \infty} \tan^{-1}(x) \rfloor = 1 \]

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**(ii) Calculate \( \lim_{x \to -\infty} \tan^{-1}(x) \)**

1. **Understanding the Function**: Again, we are dealing with the inverse tangent function \( \tan^{-1}(x) \).

2. **Behavior as \( x \to -\infty \)**: As \( x \) decreases towards negative infinity, the value of \( \tan^{-1}(x) \) approaches \( -\frac{\pi}{2} \). This is because the tangent function approaches negative infinity as its input approaches \( -\frac{\pi}{2} \).

3. **Conclusion**: Therefore, we can write:
\[ \lim_{x \to -\infty} \tan^{-1}(x) = -\frac{\pi}{2} \]

4. **Applying the Greatest Integer Function**: Since we need the greatest integer function of this limit:
\[ \lfloor \lim_{x \to -\infty} \tan^{-1}(x) \rfloor = \lfloor -\frac{\pi}{2} \rfloor \]
The value of \( -\frac{\pi}{2} \) is approximately \( -1.57 \), so:
\[ \lfloor -\frac{\pi}{2} \rfloor = -2 \]

**Final Answer for (ii)**:
\[ \lfloor \lim_{x \to -\infty} \tan^{-1}(x) \rfloor = -2 \]

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### Summary of Results:
- (i) \( \lfloor \lim_{x \to \infty} \tan^{-1}(x) \rfloor = 1 \)
- (ii) \( \lfloor \lim_{x \to -\infty} \tan^{-1}(x) \rfloor = -2 \)