Evaluate `lim_(xto2^(+)) ([x-2])/(log(x-2)),` where `[.]` represents the greatest integer function.

Explanation

To evaluate the limit

\[ \lim_{x \to 2^{+}} \frac{[x-2]}{\log(x-2)}, \]

where \([.]\) represents the greatest integer function, we will follow these steps:

### Step 1: Analyze the expression as \(x\) approaches \(2^{+}\)

As \(x\) approaches \(2\) from the right (i.e., \(x \to 2^{+}\)), the term \(x - 2\) approaches \(0^{+}\) (a small positive number).

### Step 2: Evaluate the numerator

The numerator \([x - 2]\) represents the greatest integer less than or equal to \(x - 2\). Since \(x\) is slightly greater than \(2\), \(x - 2\) will be a small positive number, which means:

\[ [x - 2] = 0 \quad \text{for } x \to 2^{+}. \]

### Step 3: Evaluate the denominator

The denominator is \(\log(x - 2)\). As \(x\) approaches \(2^{+}\), \(x - 2\) approaches \(0^{+}\), and thus:

\[ \log(x - 2) \to \log(0^{+}) \to -\infty. \]

### Step 4: Substitute into the limit

Now substituting these values into the limit gives:

\[ \lim_{x \to 2^{+}} \frac{[x - 2]}{\log(x - 2)} = \lim_{x \to 2^{+}} \frac{0}{-\infty}. \]

### Step 5: Evaluate the limit

The expression \(\frac{0}{-\infty}\) simplifies to \(0\). Therefore, we conclude that:

\[ \lim_{x \to 2^{+}} \frac{[x - 2]}{\log(x - 2)} = 0. \]

### Final Answer

Thus, the final answer is:

\[ \boxed{0}. \]