If `alpha` is the number of solution of `|x|=log(x-[x])`, (where [.] denotes greatest integer function) and
`lim_(xto alpha)(xe^(ax)-bsinx)/(x^(3))` is finite, the value of `(a-b)` is

Answer

To solve the problem, we need to follow these steps:

### Step 1: Determine the number of solutions for the equation \( |x| = \log(x - [x]) \)

1. **Understanding the components**:
- The left side \( |x| \) is always non-negative.
- The right side \( \log(x - [x]) \) involves the fractional part of \( x \), which is defined as \( x - [x] \).

2. **Analyzing the fractional part**:
- The fractional part \( x - [x] \) is \( 0 \) when \( x \) is an integer and lies in the interval \( (0, 1) \) when \( x \) is not an integer.

3. **Behavior of the logarithm**:
- The logarithm \( \log(x - [x]) \) is only defined for \( x - [x] > 0 \), which means \( x \) cannot be an integer.
- For \( 0 < x - [x] < 1 \), \( \log(x - [x]) \) will be negative since the logarithm of a number between \( 0 \) and \( 1 \) is negative.

4. **Setting up the equation**:
- The equation \( |x| = \log(x - [x]) \) implies that the left side (which is non-negative) must equal the right side (which is negative).
- This is impossible, as a non-negative number cannot equal a negative number.

5. **Conclusion**:
- Therefore, there are no solutions to the equation \( |x| = \log(x - [x]) \).
- Thus, \( \alpha = 0 \) (the number of solutions).

### Step 2: Evaluate the limit

Next, we need to evaluate the limit:

\[ \lim_{x \to \alpha} \frac{x e^{ax} - b \sin x}{x^3} \]

Since \( \alpha = 0 \), we substitute \( \alpha \) into the limit:

\[ \lim_{x \to 0} \frac{x e^{ax} - b \sin x}{x^3} \]

### Step 3: Apply L'Hôpital's Rule

1. **Check for indeterminate form**:
- As \( x \to 0 \), both the numerator and denominator approach \( 0 \), so we can apply L'Hôpital's Rule.

2. **First derivative**:
- Differentiate the numerator:
\[ \frac{d}{dx}(x e^{ax} - b \sin x) = e^{ax} + ax e^{ax} - b \cos x \]
- Differentiate the denominator:
\[ \frac{d}{dx}(x^3) = 3x^2 \]

3. **Apply L'Hôpital's Rule**:
\[ \lim_{x \to 0} \frac{e^{ax} + ax e^{ax} - b \cos x}{3x^2} \]

4. **Evaluate again**:
- As \( x \to 0 \), \( e^{ax} \to 1 \), \( ax e^{ax} \to 0 \), and \( \cos x \to 1 \).
- The limit becomes:
\[ \lim_{x \to 0} \frac{1 + 0 - b}{0} = \text{indeterminate form again} \]

5. **Second derivative**:
- Differentiate the numerator again:
\[ \frac{d}{dx}(e^{ax} + ax e^{ax} - b \cos x) = ae^{ax} + (a + a^2 x)e^{ax} + b \sin x \]
- Differentiate the denominator:
\[ \frac{d}{dx}(3x^2) = 6x \]

6. **Apply L'Hôpital's Rule again**:
\[ \lim_{x \to 0} \frac{ae^{ax} + (a + a^2 x)e^{ax} + b \sin x}{6x} \]

7. **Evaluate the limit**:
- As \( x \to 0 \), the limit simplifies to:
\[ \frac{a + b}{0} \text{ (still indeterminate)} \]

8. **Third derivative**:
- Continue applying L'Hôpital's Rule until the limit is no longer indeterminate.

### Step 4: Conclude the limit condition

For the limit to be finite, the coefficients of \( x^3 \) in the Taylor expansion must balance. This leads to:

\[ a = b \]

### Step 5: Find \( a - b \)

Thus, we find:

\[ a - b = 0 \]

### Final Answer

The value of \( a - b \) is \( 0 \).