If ` m,n in I^(+)`, then `lim_(x to 0) ("sin"x^(n))/(("sin"x)^(m))` equals

Options

A. 1, if `n lt m`
B. 0, if n = m
C. `(n)/(m)`
D. 0, if `n gt m`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m} \), we will use the known limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and analyze the behavior of the function as \( x \) approaches 0.

### Step-by-Step Solution:

1. **Rewrite the limit**:
We start with the expression:
\[ \lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m} \]

2. **Use the limit property**:
We know that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Therefore, we can express \( \sin(x^n) \) and \( \sin x \) in terms of \( x \):
\[ \sin(x^n) \approx x^n \quad \text{as } x \to 0 \]
\[ \sin x \approx x \quad \text{as } x \to 0 \]

3. **Substituting the approximations**:
Substitute these approximations into the limit:
\[ \lim_{x \to 0} \frac{x^n}{(x)^m} = \lim_{x \to 0} \frac{x^n}{x^m} = \lim_{x \to 0} x^{n-m} \]

4. **Analyze the limit based on \( n \) and \( m \)**:
Now we analyze the limit based on the relationship between \( n \) and \( m \):
- **Case 1**: If \( n = m \):
\[ \lim_{x \to 0} x^{n-m} = \lim_{x \to 0} x^0 = 1 \]
- **Case 2**: If \( n > m \):
\[ \lim_{x \to 0} x^{n-m} = \lim_{x \to 0} x^{\text{positive}} = 0 \]
- **Case 3**: If \( n < m \):
\[ \lim_{x \to 0} x^{n-m} = \lim_{x \to 0} \frac{1}{x^{\text{positive}}} = \infty \]

5. **Conclusion**:
Therefore, we can summarize the results:
\[ \lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m} = \begin{cases} 1 & \text{if } n = m \\ 0 & \text{if } n > m \\ \infty & \text{if } n < m \end{cases} \]