`lim_(xrarr0) [(100 tan x sin x)/(x^2)]` is (where `[.]` represents greatest integer function).

Options

A. 99
B. 100
C. 0
D. non-existent

Explanation

To solve the limit \( \lim_{x \to 0} \frac{100 \tan x \sin x}{x^2} \) and find the greatest integer function of the result, we can follow these steps:

### Step 1: Analyze the limit
We start with the expression:
\[ \lim_{x \to 0} \frac{100 \tan x \sin x}{x^2} \]

### Step 2: Use known limits
We know that as \( x \) approaches 0:
- \( \tan x \approx x \) (more precisely, \( \tan x \to x \) as \( x \to 0 \))
- \( \sin x \approx x \) (more precisely, \( \sin x \to x \) as \( x \to 0 \))

### Step 3: Substitute the approximations
Using these approximations, we can rewrite the limit:
\[ \tan x \sin x \approx x \cdot x = x^2 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{100 \tan x \sin x}{x^2} \approx \lim_{x \to 0} \frac{100 \cdot x^2}{x^2} \]

### Step 4: Simplify the expression
This simplifies to:
\[ \lim_{x \to 0} 100 = 100 \]

### Step 5: Consider the behavior near zero
However, we need to be careful about the behavior of \( \tan x \) and \( \sin x \) as \( x \) approaches 0. We know:
- \( \tan x = \frac{\sin x}{\cos x} \)
- As \( x \to 0 \), \( \cos x \to 1 \)

Thus:
\[ \tan x \sin x = \frac{\sin^2 x}{\cos x} \]
And we can rewrite our limit as:
\[ \lim_{x \to 0} \frac{100 \sin^2 x}{x^2 \cos x} \]

### Step 6: Use the limit of \( \sin^2 x \)
Using the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):
\[ \lim_{x \to 0} \frac{\sin^2 x}{x^2} = 1 \]
Thus:
\[ \lim_{x \to 0} \frac{100 \sin^2 x}{x^2 \cos x} = \frac{100 \cdot 1}{1} = 100 \]

### Step 7: Final result
Since the limit approaches 100, we consider the greatest integer function:
\[ \lfloor 100 \rfloor = 99 \]

### Conclusion
Thus, the final answer is:
\[ \boxed{99} \]