`lim_(xto0) (x^(a)sin^(b)x)/(sin(x^(c)))`, where `a,b,c inR~{0},` exists and has non-zero value. Then,

Options

A. 0
B. `pi//2`
C. `pi`
D. `2pi`

Explanation

To solve the limit problem \( \lim_{x \to 0} \frac{x^a \sin^b x}{\sin(x^c)} \), where \( a, b, c \in \mathbb{R} \setminus \{0\} \), we need to analyze the behavior of the function as \( x \) approaches 0.

### Step-by-step Solution:

1. **Rewrite the Limit**:
We start by rewriting the limit expression:
\[ \lim_{x \to 0} \frac{x^a \sin^b x}{\sin(x^c)} = \lim_{x \to 0} \frac{x^a (\sin x)^b}{\sin(x^c)} \]

2. **Use the Sine Limit Property**:
We know that as \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \). Therefore, we can rewrite \( \sin x \) as \( x \) times a limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \sin x \approx x \text{ for small } x \]
Thus, we can express \( \sin^b x \) as:
\[ \sin^b x = (x \cdot \frac{\sin x}{x})^b = x^b \left( \frac{\sin x}{x} \right)^b \]

3. **Rewrite the Expression**:
Substitute this back into our limit:
\[ \lim_{x \to 0} \frac{x^a (x^b \left( \frac{\sin x}{x} \right)^b)}{\sin(x^c)} = \lim_{x \to 0} \frac{x^{a+b} \left( \frac{\sin x}{x} \right)^b}{\sin(x^c)} \]

4. **Analyze the Denominator**:
For small \( x \), \( x^c \) also approaches 0, so we can use the same sine limit property:
\[ \sin(x^c) \approx x^c \]
Thus, we can rewrite our limit as:
\[ \lim_{x \to 0} \frac{x^{a+b} \left( \frac{\sin x}{x} \right)^b}{x^c} \]

5. **Simplify the Limit**:
This simplifies to:
\[ \lim_{x \to 0} x^{a+b-c} \left( \frac{\sin x}{x} \right)^b \]

6. **Evaluate the Limit**:
As \( x \to 0 \), \( \left( \frac{\sin x}{x} \right)^b \to 1 \). Therefore, we focus on the term \( x^{a+b-c} \):
- For the limit to exist and be non-zero, the exponent \( a + b - c \) must equal 0.
- This gives us the condition:
\[ a + b - c = 0 \implies a + b = c \]

### Conclusion:
Thus, the value of \( a + b \) is equal to \( c \).