The value of `lim_(x to 0) (["11x"/"sinx"]+["21 sinx"/x])`, where [x] is the greatest integer less than or equal to x is

Options

A. 32
B. 31
C. 11
D. 21

Answer

To solve the limit \( \lim_{x \to 0} \left( \left\lfloor \frac{11x}{\sin x} \right\rfloor + \left\lfloor \frac{21 \sin x}{x} \right\rfloor \right) \), we will analyze each term separately.

### Step 1: Analyze \( \frac{11x}{\sin x} \)

As \( x \to 0 \), we know from the limit property that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \lim_{x \to 0} \frac{x}{\sin x} = 1 \]
Thus,
\[ \lim_{x \to 0} \frac{11x}{\sin x} = 11 \cdot \lim_{x \to 0} \frac{x}{\sin x} = 11 \cdot 1 = 11 \]

### Step 2: Determine \( \left\lfloor \frac{11x}{\sin x} \right\rfloor \)

As \( x \) approaches 0, \( \frac{11x}{\sin x} \) approaches 11. However, we need to consider the behavior of \( \frac{11x}{\sin x} \) as \( x \) approaches 0 from both sides.

- For small positive \( x \), \( \sin x < x \), hence \( \frac{11x}{\sin x} > 11 \).
- For small negative \( x \), \( \sin x > x \), hence \( \frac{11x}{\sin x} < 11 \).

Thus, as \( x \to 0 \), \( \left\lfloor \frac{11x}{\sin x} \right\rfloor \) will equal 10 for both sides.

### Step 3: Analyze \( \frac{21 \sin x}{x} \)

Similarly, we know:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \lim_{x \to 0} \frac{21 \sin x}{x} = 21 \]

### Step 4: Determine \( \left\lfloor \frac{21 \sin x}{x} \right\rfloor \)

As \( x \) approaches 0, \( \frac{21 \sin x}{x} \) approaches 21.

- For small positive \( x \), \( \sin x < x \), hence \( \frac{21 \sin x}{x} < 21 \).
- For small negative \( x \), \( \sin x > x \), hence \( \frac{21 \sin x}{x} > 21 \).

Thus, as \( x \to 0 \), \( \left\lfloor \frac{21 \sin x}{x} \right\rfloor \) will equal 20 for both sides.

### Step 5: Combine the results

Now we can combine the results:
\[ \lim_{x \to 0} \left( \left\lfloor \frac{11x}{\sin x} \right\rfloor + \left\lfloor \frac{21 \sin x}{x} \right\rfloor \right) = 10 + 20 = 30 \]

### Final Answer

Thus, the value of the limit is:
\[ \boxed{30} \]