Prove that `f(x) = [tan x] + sqrt(tan x - [tan x])`. (where [.] denotes greatest integer function) is continuous in `[0, (pi)/(2))`.

Answer

To prove that the function \( f(x) = [\tan x] + \sqrt{\tan x - [\tan x]} \) is continuous in the interval \( [0, \frac{\pi}{2}) \), we will analyze the function step by step.

### Step 1: Understanding the Components of the Function
The function consists of two parts:
1. The greatest integer function \( [\tan x] \), which gives the largest integer less than or equal to \( \tan x \).
2. The term \( \sqrt{\tan x - [\tan x]} \), which represents the square root of the fractional part of \( \tan x \).

### Step 2: Analyzing the Behavior of \( \tan x \)
The function \( \tan x \) is continuous and increasing in the interval \( [0, \frac{\pi}{2}) \):
- At \( x = 0 \), \( \tan 0 = 0 \).
- As \( x \) approaches \( \frac{\pi}{2} \), \( \tan x \) approaches \( +\infty \).

### Step 3: Identifying the Range of \( \tan x \)
In the interval \( [0, \frac{\pi}{2}) \):
- For \( 0 \leq x < \frac{\pi}{4} \), \( 0 \leq \tan x < 1 \) implies \( [\tan x] = 0 \).
- For \( \frac{\pi}{4} \leq x < \tan^{-1}(2) \), \( 1 \leq \tan x < 2 \) implies \( [\tan x] = 1 \).
- For \( \tan^{-1}(2) \leq x < \tan^{-1}(3) \), \( 2 \leq \tan x < 3 \) implies \( [\tan x] = 2 \).
- This pattern continues as \( x \) increases towards \( \frac{\pi}{2} \).

### Step 4: Expressing \( f(x) \) in Different Intervals
1. **For \( 0 \leq x < \frac{\pi}{4} \)**:
\[ f(x) = 0 + \sqrt{\tan x} = \sqrt{\tan x} \]

2. **For \( \frac{\pi}{4} \leq x < \tan^{-1}(2) \)**:
\[ f(x) = 1 + \sqrt{\tan x - 1} \]

3. **For \( \tan^{-1}(2) \leq x < \tan^{-1}(3) \)**:
\[ f(x) = 2 + \sqrt{\tan x - 2} \]

### Step 5: Continuity at the Transition Points
To prove continuity, we need to check the limits at the transition points:
- **At \( x = \frac{\pi}{4} \)**:
\[ \lim_{x \to \frac{\pi}{4}^-} f(x) = \sqrt{\tan(\frac{\pi}{4})} = \sqrt{1} = 1 \]
\[ \lim_{x \to \frac{\pi}{4}^+} f(x) = 1 + \sqrt{\tan(\frac{\pi}{4}) - 1} = 1 + 0 = 1 \]
Thus, \( f(x) \) is continuous at \( x = \frac{\pi}{4} \).

- **At \( x = \tan^{-1}(2) \)**:
\[ \lim_{x \to \tan^{-1}(2)^-} f(x) = 1 + \sqrt{2 - 1} = 1 + 1 = 2 \]
\[ \lim_{x \to \tan^{-1}(2)^+} f(x) = 2 + \sqrt{\tan(\tan^{-1}(2)) - 2} = 2 + 0 = 2 \]
Thus, \( f(x) \) is continuous at \( x = \tan^{-1}(2) \).

### Conclusion
Since \( f(x) \) is continuous in each sub-interval and at the transition points, we conclude that \( f(x) \) is continuous in the entire interval \( [0, \frac{\pi}{2}) \).

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