Find ` lim_( x to 0) (sin x^(n))/((sin x)^(m)) " where" , m , n in Z^(+)` equal

Options

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

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m} \) where \( m, n \in \mathbb{Z}^+ \), we will follow these steps:

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

### Step 2: Use the small angle approximation
Recall that for small values of \( x \), \( \sin x \approx x \). Therefore, we can use this approximation:
\[ \sin(x^n) \approx x^n \quad \text{and} \quad \sin x \approx x \]

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

### Step 4: Analyze the limit based on the value of \( n - m \)
Now we analyze the limit based on the exponent \( n - m \):
1. **If \( n > m \)**:
\[ \lim_{x \to 0} x^{n-m} = 0 \]
2. **If \( n = m \)**:
\[ \lim_{x \to 0} x^{n-m} = \lim_{x \to 0} x^0 = 1 \]
3. **If \( n < m \)**:
\[ \lim_{x \to 0} x^{n-m} = \lim_{x \to 0} \frac{1}{x^{m-n}} = \infty \]

### Conclusion
Thus, the limit can be summarized as follows:
- If \( n > m \), the limit is \( 0 \).
- If \( n = m \), the limit is \( 1 \).
- If \( n < m \), the limit is \( \infty \).

### Final Answer
The limit \( \lim_{x \to 0} \frac{\sin(x^n)}{(\sin x)^m} \) equals:
- \( 0 \) if \( n > m \)
- \( 1 \) if \( n = m \)
- \( \infty \) if \( n < m \)

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