Eavaluate `lim_(x to 0) (sin^(2)2x)/(sin^(2)4x)`

Options

A. `1/2`
B. `1`
C. `1/4`
D. `2`

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} \), we can follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit expression:
\[ \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} = \lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \cdot \frac{2x}{4x} \cdot \frac{4x}{\sin(4x)} \right)^2 \]

### Step 2: Split the limit
We can split the limit into three parts:
\[ = \lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \right)^2 \cdot \left( \frac{2x}{4x} \right)^2 \cdot \lim_{x \to 0} \left( \frac{4x}{\sin(4x)} \right)^2 \]

### Step 3: Evaluate each limit
1. **First limit**:
\[ \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \]
Therefore, \( \left( \lim_{x \to 0} \frac{\sin(2x)}{2x} \right)^2 = 1^2 = 1 \).

2. **Second limit**:
\[ \frac{2x}{4x} = \frac{1}{2} \quad \text{(this is constant)} \]
Therefore, \( \left( \frac{1}{2} \right)^2 = \frac{1}{4} \).

3. **Third limit**:
\[ \lim_{x \to 0} \frac{4x}{\sin(4x)} = 1 \]
Therefore, \( \left( \lim_{x \to 0} \frac{4x}{\sin(4x)} \right)^2 = 1^2 = 1 \).

### Step 4: Combine the results
Now, we combine all the results:
\[ \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} = 1 \cdot \frac{1}{4} \cdot 1 = \frac{1}{4} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin^2(2x)}{\sin^2(4x)} = \frac{1}{4} \]