The value of `lim_(xto 0)[(sinx.tanx)/(x^(2))]` is ……..
(where [.] denotes greatest integer function).

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin x \tan x}{x^2} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to 0} \frac{\sin x \tan x}{x^2} \]
Recall that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, we can rewrite the expression as:
\[ \lim_{x \to 0} \frac{\sin x \cdot \frac{\sin x}{\cos x}}{x^2} = \lim_{x \to 0} \frac{\sin^2 x}{x^2 \cos x} \]

### Step 2: Split the limit
We can separate the limit into two parts:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} \cdot \lim_{x \to 0} \frac{1}{\cos x} \]

### Step 3: Evaluate the first limit
We know from standard limits that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right)^2 = 1^2 = 1 \]

### Step 4: Evaluate the second limit
Next, we evaluate:
\[ \lim_{x \to 0} \frac{1}{\cos x} \]
Since \( \cos(0) = 1 \), we have:
\[ \lim_{x \to 0} \frac{1}{\cos x} = \frac{1}{1} = 1 \]

### Step 5: Combine the results
Now we combine the results from the two limits:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} \cdot \lim_{x \to 0} \frac{1}{\cos x} = 1 \cdot 1 = 1 \]

### Step 6: Apply the greatest integer function
The question asks for the greatest integer function of the limit:
\[ \lfloor 1 \rfloor = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]